L.C. MATHS

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Past Papers

## Statistics

NOTE: Clicking an entry in the right column will take you directly to that question!

(BLUE = Paper 1, RED = Paper 2)

Since 2015, the following subtopic... Has explicitly appeared in...

Mean, Mode & Median

Stem and Leaf Diagrams

Has not appeared

Histograms

Has not appeared

Scatter Diagrams

Normal Distributions

Inferential Statistics (Proportion)

Inferential Statistics (Mean)

## 2022 Paper 2 Question 5

(a) A survey on remote learning was carried out on a random sample of $$400$$ students.
$$135$$ of the students preferred remote learning over in-person learning.
For parts (a)(i), (a)(ii), and (a)(iii), give all solutions as decimals, correct to $$4$$ decimal places.

(i) Work out the proportion of the sample that preferred remote learning.

(ii) Use the margin of error $$\left(\frac{1}{\sqrt{n}}\right)$$ to create a $$95\%$$ confidence interval for the proportion of the population that preferred remote learning.

(iii) Using the proportion from part (a)(i), create a $$95\%$$ confidence interval for this population proportion that is more accurate than the $$95\%$$ confidence interval based on the margin of error.

(i) $$0.3375$$

(ii) $$[0.2875,0.3875]$$

(iii) $$[0.2912,0.3838]$$

Solution

(i)

\begin{align}\hat{p}&=\frac{135}{400}\\&=0.3375\end{align}

(ii)

\begin{align}\hat{p}\pm\mbox{error }&=0.3375\pm\frac{1}{\sqrt{400}}\\&=0.3375\pm0.05\end{align}

\begin{align}\downarrow\end{align}

Confidence Interval: $$[0.2875,0.3875]$$

(iii)

\begin{align}\hat{p}\pm 1.96\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}&=0.3375\pm1.96\sqrt{\frac{(0.3375)(1-0.3375)}{400}}\\&=0.3375\pm0.04633…\end{align}

\begin{align}\downarrow\end{align}

Confidence Interval: $$[0.2912,0.3838]$$

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(b) In $$2019$$, people with a pre-pay mobile phone plan spent an average (mean) of €$$20.79$$ on
their mobile phone each month (source: www.comreg.ie).

In $$2021$$, some students carried out a survey to see if this figure had changed.
They surveyed a random sample of $$500$$ people with pre-pay mobile phone plans.
For this sample, the mean amount spent per month was €$$22.16$$ and the standard deviation was €$$8.12$$.

Carry out a hypothesis test at the $$5\%$$ level of significance to see if this shows a change in the mean monthly spend on mobile phones for people with a pre-pay plan.

The average amount has changed.

Solution

Null Hypothesis: Average amount has not changed.

Alternative Hypothesis: Average amount has changed.

$\,$

Calculations:

\begin{align}\hat{p}\pm 1.96\frac{\sigma}{\sqrt{n}}&=22.16\pm1.96\frac{8.12}{\sqrt{500}}\\&=22.16\pm0.7117…\end{align}

\begin{align}\downarrow\end{align}

Confidence Interval: $$[21.44…,22.87…]$$

$\,$

Conclusion: As $$20.79$$ lies outside of this interval, the average amount has changed.

Video Walkthrough

## 2022 Paper 2 Question 8(a)-(c)

(a) Jena is researching fuel consumption in cars. She finds the following data for the number of miles per gallon ($$\mbox{m/g}$$) for eight different cars, labelled A to H, when driving in the city and on the motorway:

Car City ($$\mbox{m/g}$$) Motorway ($$\mbox{m/g}$$)

A

$$22$$

$$34$$

B

$$27$$

$$38$$

C

$$24$$

$$34$$

D

$$16$$

$$27$$

E

$$15$$

$$24$$

F

$$21$$

$$30$$

G

$$\mathbf{30}$$

$$\mathbf{40}$$

H

$$\mathbf{17}$$

$$\mathbf{30}$$

(i) The scatterplot below shows this data for cars A to F.

Using the data in the table above, plot and label points to represent cars G and H on the scatterplot below.

(ii) On the scatterplot, draw the line of best fit for the data, by eye.

(iii) Two other cars, K and L, have the miles per gallon values given in the following table.

Use your line of best fit on the scatterplot to fill in an estimate for each of the two missing values in the table below. Show your work on the scatterplot.

Car City ($$\mbox{m/g}$$) Motorway ($$\mbox{m/g}$$)

K

$$20$$

L

$$60$$

(iv) Based on the data given, would you be more confident in the value you estimated for K or for L? Give a reason for your answer.

(v) Find the value of $$r$$, the correlation coefficient between city and motorway miles per gallon. Use only the values for the $$8$$ cars A to H in the table on the previous page.
Give your answer correct to $$3$$ decimal places.

(i)

(ii)

(iii)

Car City ($$\mbox{m/g}$$) Motorway ($$\mbox{m/g}$$)

K

$$20$$

$$32$$

L

$$56$$

$$60$$

(iv) $$K$$ as it is in the neighbourhood of the other points.

(v) $$r\approx0.966$$

Solution

(i)

(ii)

(iii)

Car City ($$\mbox{m/g}$$) Motorway ($$\mbox{m/g}$$)

K

$$20$$

$$32$$

L

$$56$$

$$60$$

(iv) $$K$$ as it is in the neighbourhood of the other points.

(v) Calculator: $$r\approx0.966$$

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(b) The scatterplot shows some values of fuel consumption ($$F$$) for the given values of engine speed ($$S$$), for a particular car.
For the points in this scatterplot, $$F$$ can be closely approximated by a quadratic function of $$S$$.

$$r_{FS}$$ is the correlation coefficient between $$F$$ and $$S$$, based on the points in this scatterplot.
Give a reason why you might think that $$r_{FS}$$ is very close to $$0$$.

The line of best fit is close to being horizontal.

Solution

The line of best fit is close to being horizontal.

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(c) $$13$$ customers rated their experience in a garage, by giving a whole-number score out of $$100$$.
The mean score was $$52$$. The median score was $$54$$. No two scores were the same.

The data below shows the score for each of the $$13$$ customers (in no particular order).
Stephen gave a score of S and Mary gave a score of M, where S, M $$\in\mathbb{N}$$.
Find the least value and the greatest value that S could be.

\begin{align}46&&68&&24&&74&&42&&30&&61&&54&&28&&50&&57&&\mbox{S}&&\mbox{M}\end{align}

Least Value: $$55$$

Greatest Value: $$87$$

Solution

Least Value

The $$11$$ know numbers listed in order are

\begin{align}24&&28&&30&&42&&46&&50&&54&&57&&61&&68&&74\end{align}

The median of these $$11$$ numbers is therefore $$50$$.

The only way the median can increase to $$54$$ is if both $$S$$ and $$M$$ are larger than $$54$$.

Therefore, the smallest possible value of $$S$$ is $$55$$.

$\,$

Greatest Value

\begin{align}\frac{46+68+24+74+42+30+61+54+28+50+57+S+M}{13}=52\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{534+S+M}{13}=52\end{align}

\begin{align}\downarrow\end{align}

\begin{align}534+S+M=676\end{align}

\begin{align}\downarrow\end{align}

\begin{align}S=142-M\end{align}

The smallest possible value of $$M$$ is $$55$$. Therefore, the largest possible value of $$S$$ is

\begin{align}S&=142-M\\&=142-55\\&=87\end{align}

Video Walkthrough

## 2022 Paper 2 Question 10(a) & (e)

(a) In an athletics competition, there were a number of heats of the $$1500\mbox{ m}$$ race.
In the heats, the times that it took the runners to complete the $$1500\mbox{ m}$$ were approximately normally distributed, with a mean time of $$225$$ seconds and a standard deviation of $$12$$ seconds.

(i) Find the percentage of runners in these heats who took more than $$240$$ seconds to run the $$1500\mbox{ m}$$.

(ii) The $$20\%$$ of runners with the fastest times qualified for the final.
Assuming the race times were normally distributed as described above, work out the time needed to qualify for the final, correct to the nearest second.

(i) $$10.56\%$$

(ii) $$215\mbox{ s}$$

Solution

(i)

\begin{align}z&=\frac{x-\mu}{\sigma}\\&=\frac{240-225}{12}\\&=1.25\end{align}

\begin{align}\downarrow\end{align}

\begin{align}P(x>240)&=P(z>1.25)\\&=1-P(z<1.25)\\&=1-0.8944\\&=0.1056\end{align}

Therefore, the percentage of runners in these heats who took more than $$240$$ seconds to run the $$1500\mbox{ m}$$ is $$10.56\%$$.

(ii) When $$P=0.8$$, $$z\approx0.84$$ and therefore

\begin{align}0.84=\frac{225-x}{12}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x&=225-12(0.84)\\&\approx215\mbox{ s}\end{align}

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(e) Sorcha ran two different marathons: the Windy Marathon and the Sunny Marathon.
The table below gives some details on the finishing times for the two marathons.
For each marathon, the finishing times of the runners were approximately normally distributed.

Sorcha’s position in each marathon was based on her finishing time.
Sorcha came $$5265$$th in the Windy Marathon, so exactly $$5264$$ of the $$6000$$ runners had a finishing time that was less than Sorcha’s.

Sorcha’s finishing time for both marathons was the same.

Use this fact, and the details in the table, to estimate Sorcha’s position in the Sunny Marathon. Show all of your working out.

Mean finishing time (minutes) Standard deviation of finishing times (minutes) Number of runners Sorcha's position

Windy Marathon

$$254$$

$$38$$

$$6000$$

$$5265$$th

Sunny Marathon

$$247$$

$$29$$

$$2000$$

$$1{,}923$$

Solution

Windy Marathon

\begin{align}P=\frac{5{,}265}{6{,}000}=0.8775\end{align}

\begin{align}\downarrow\end{align}

\begin{align}z\approx1.16\end{align}

\begin{align}\downarrow\end{align}

\begin{align}1.16&=\frac{x-\mu}{\sigma}\\&=\frac{t-254}{38}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}t&=1.16(38)+254\\&=298.08\mbox{ min}\end{align}

$\,$

Sunny Marathon

\begin{align}z&=\frac{x-\mu}{\sigma}\\&=\frac{298.08-247}{29}\\&\approx1.76\end{align}

\begin{align}\downarrow\end{align}

\begin{align}P=0.9608\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{\mbox{Position}}{2{,}000}=0.9608\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\mbox{Position}&=(0.9608)(2{,}000)\\&\approx1{,}923\end{align}

Video Walkthrough

## 2021 Paper 2 Question 8(a)-(b)

(a) In a school all First Years sat a common maths exam.
The results, in integer values, were normally distributed with a mean of $$176$$ marks and a standard deviation of $$36$$ marks.
The top $$10\%$$ of students will go forward to a county maths competition.

(i) Find the minimum mark needed on the exam to progress to the county stage.

(ii) The school awarded a Certificate of Merit to any student who achieved between $$165$$ marks and $$210$$ marks.
Find the percentage of First Years who received the Certificate of Merit

$$223$$

(ii) $$44.81\%$$

Solution

(i)

\begin{align}P(z<1.28)&=0.8997\end{align}

\begin{align}z=\frac{x-\mu}{\sigma}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}1.28=\frac{x-176}{36}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x&=(1.28)(36)+176\\&=222.08\end{align}

Therefore, the minimum mark is $$223$$.

(ii)

\begin{align}P\left(z<\frac{210-176}{36}\right)-P\left(z>\frac{165-176}{36}\right)&=P(z<0.94)-P(z>-0.31)\\&=0.8264-(1-0.6217)\\&=0.4481\end{align}

Therefore, $$44.81\%$$ of students got the Certificate of Merit.

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(b) A news report claimed that 6th year students in Ireland studied an average of $$21$$ hours per week, outside of class time. A Leaving Cert class surveyed $$60$$ students in 6th year, chosen at random, from different schools. It found that the average study time was $$19.8$$ hours and the standard deviation was $$5.2$$ hours.

(i) Find the test statistic (the $$z$$-score) of this sample mean.

(ii) Find the $$p$$-value of this test statistic. Comment on what can be concluded from its value, in a two-tailed hypothesis test at the $$5\%$$ level of significance, in relation to the news report claim.

(i) $$-1.787…$$

(ii) $$p=0.0734$$. As $$p>0.05$$, we do not have enough evidence to state that the claim is incorrect.

Solution

(i)

\begin{align}z&=\frac{\bar{x}-\mu}{\sigma/\sqrt{n}}\\&=\frac{19.8-21}{5.2/\sqrt{60}}\\&=-1.787…\end{align}

(ii)

\begin{align}p&=2[1-P(z<1.79)]\\&=2(1-0.9633)\\&=0.0734\end{align}

As $$0.0734>0.05$$, we do not have enough evidence to state that the claim is incorrect.

Video Walkthrough

## 2020 Paper 2 Question 8

(a) An airline company Trans-sky Airways has designed an aptitude test for people applying for jobs as trainee pilots. The aptitude test is scored out of $$500$$ marks. The results are normally distributed with a mean score of $$280$$ and a standard deviation of $$90$$.

(i) The top $$25\%$$ of people taking the aptitude test are invited back for an interview.
Find the minimum mark needed on the test in order to be invited back for interview.

(ii) Anyone who scores above the $$40$$th percentile can re-sit the test later.
Eileen scored $$260$$ marks in the test.
Find out whether or not Eileen is eligible to re-sit the test.

(i) $$342$$

(ii) Eileen is eligible to re-sit the test.

Solution

(i)

\begin{align}z=\frac{x-\mu}{\sigma}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}0.68=\frac{x-280}{90}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x&=280+(0.68)(90)\\&=341.2\end{align}

Therefore, the minimum mark is $$342$$.

(ii)

\begin{align}z&=\frac{260-280}{90}\\&=-0.222\end{align}

Since $$z=-0.25$$ corresponds to the 40th percentile, Eileen is eligible to re-sit the test.

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(b)

(i) Explain the relevance of the z-scores $$-1.96$$ and $$1.96$$ in the standard normal distribution.

(ii) Trans-sky Airways surveyed $$2500$$ of its passengers about a new service it proposed to introduce. The variable $$\hat{p}$$ is the proportion of respondents in the survey who said they would use the new service.
The radius of the $$95\%$$ confidence interval of the survey was $$0.01568$$.
Find the value of $$\hat{p}$$, where $$0.5<\hat{p}\leq1$$.

(i) $$95\%$$ of the curve is found in the interval $$-1.96\leq z\leq1.96$$.

(ii) $$\hat{p}=0.8$$

Solution

(i) $$95\%$$ of the curve is found in the interval $$-1.96\leq z\leq1.96$$

(ii)

\begin{align}1.96\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}=r\end{align}

\begin{align}\downarrow\end{align}

\begin{align}1.96\sqrt{\frac{\hat{p}(1-\hat{p})}{2500}}=0.01568\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\hat{p}(1-\hat{p})=2500\left(\frac{0.01568^2}{1.96^2}\right)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\hat{p}^2-\hat{p}+0.16=0\end{align}

\begin{align}\downarrow\end{align}

$$\hat{p}=0.2$$ or $$\hat{p}=0.8$$

\begin{align}\downarrow\end{align}

$$\hat{p}=0.8$$

(since $$0.5<\hat{p}\leq1$$).

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(c) The weight of the Airline passengers’ carry-on luggage is normally distributed with a mean of $$12\mbox{ kg}$$. The Airline has recently introduced a fee for non-carry-on luggage. After the fee was introduced, the Airline expected the mean weight of the carry-on luggage to change.
They selected a random sample of $$80$$ passengers and weighed their carry-on luggage.
The sample mean was $$13.1\mbox{ kg}$$ and the sample standard deviation was $$4.5\mbox{ kg}$$.

Test the hypothesis, at the $$5\%$$ level of significance, that the mean weight of the carry-on luggage has changed. State the null hypothesis and the alternative hypothesis.
Give your conclusion in the context of the question.

The mean weight of the carry-on luggage has changed.

Solution

Null hypothesis: Mean weight of the carry-on luggage has not changed.

Alternative hypothesis: Mean weight of the carry-on luggage has changed.

Calculations:

\begin{align}z&=\frac{\bar{x}-\mu}{\sigma/\sqrt{n}}\\&=\frac{13.1-12}{4.5/\sqrt{80}}\\&=2.186…\end{align}

Conclusion

As $$z>1.96$$, the mean weight of the carry-on luggage has changed.

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(d) The company bus can carry passengers up to a total maximum weight allowance of $$3000\mbox{ kg}$$.
The weight of passengers is normally distributed with a mean of $$73\mbox{ kg}$$ and a standard deviation of $$12\mbox{ kg}$$.
$$40$$ passengers board the bus.
Find the probability that the total passenger weight will be over the maximum weight allowance.
Give your answer as a percentage correct to $$2$$ decimal places.

$$14.69\%$$

Solution

The average weight of each person at the maximum wight allowance is

\begin{align}\mu&=\frac{3000}{4}\\&=75\mbox{ kg}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}z&=\frac{75-73}{12/\sqrt{40}}\\&=1.054\end{align}

\begin{align}\downarrow\end{align}

\begin{align}P(\bar{x}<75)=0.8531\end{align}

\begin{align}\downarrow\end{align}

\begin{align}P(\bar{x}>75)&=1-P(\bar{x}<75)\\&=1-0.8531…\\&=0.1469…\end{align}

Therefore, the probability that the total passenger weight will be over the maximum weight allowance is $$14.69\%$$.

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(e) A list consists of eight whole numbers. They are labelled from $$A$$ to $$H$$ as shown below.
The numbers are all greater than zero and are ordered from smallest to largest.
The difference between any two adjacent numbers is $$2$$ or more.
The median of the list is $$12.5$$.
The lower quartile (the median of the $$4$$ lowest numbers) of the list is $$7.5$$.
The interquartile range is $$12$$.
The second largest number is $$23$$, as shown.
The range of the list is $$21$$.
The mean of the list is $$13.5$$.

Find the numbers which satisfy all of the above conditions and write them into the boxes below.

$$(4,6,9,11,14,16,23,25)$$

Solution

\begin{align}D+E&=2(12.5)\\&=25\end{align}

and

\begin{align}B+C&=2(7.5)\\&=15\end{align}

and

\begin{align}F+G&=2(12+7.5)\\&=39\end{align}

\begin{align}\downarrow\end{align}

\begin{align}F&=39-G\\&=39-23\\&=16\end{align}

and

\begin{align}H-A=21\end{align}

and

\begin{align}\frac{A+B+C+D+E+F+G+H}{8}=13.5\end{align}

\begin{align}\downarrow\end{align}

\begin{align}A+B+C+D+E+F+G+H=108\end{align}

\begin{align}\downarrow\end{align}

\begin{align}A+15+25+16+23+H=108\end{align}

\begin{align}\downarrow\end{align}

\begin{align}A+H=29\end{align}

We therefore have the following two equations for two unknowns:

\begin{align}H-A=21\end{align}

\begin{align}A+H=29\end{align}

\begin{align}\downarrow\end{align}

\begin{align}2H=50\end{align}

\begin{align}\downarrow\end{align}

\begin{align}H=25\end{align}

\begin{align}\downarrow\end{align}

\begin{align}A&=29-H\\&=29-25\\&=4\end{align}

Since $$D+E=25$$, the only possibilities are $$10+15$$, $$11+14$$ and $$12+13$$. The final option is ruled out as the difference between both numbers would be less than $$2$$. Likewise, the first option is ruled out as $$E$$ and $$F$$ would have the same problem.

Therefore, $$D=5$$ and $$E=25$$.

Likewise, since $$B+C=15$$, the only possibilities are $$5+10$$, $$6+9$$ and $$7+8$$. The final option is ruled out as the difference between both numbers would be less than $$2$$. Likewise, the first option is ruled out as $$A$$ and $$B$$ would have the same problem.

Therefore, $$B=6$$ and $$C=9$$.

Video Walkthrough

## 2019 Paper 1 Question 6(a)

(a)

(i) Given that $$x-\sqrt{32}=\sqrt{128}-5x$$, find the value of $$x$$, where $$x\in\mathbb{R}$$.
Give your answer in the form $$a\sqrt{2}$$, where $$a\in\mathbb{N}$$.

(ii) $$A=\{\sqrt{32k^2},\sqrt{50k^2},\sqrt{128k^2},\sqrt{98k^2}\}$$, where $$k\in\mathbb{N}$$.

Show that the mean of set $$A$$ is equal to the median of set $$A$$.

(i) $$2\sqrt{2}$$

Solution

(i)

\begin{align}x-\sqrt{32}=\sqrt{128}-5x\end{align}

\begin{align}\downarrow\end{align}

\begin{align}6x=\sqrt{128}+\sqrt{32}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}6x=8\sqrt{2}+4\sqrt{2}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}6x=12\sqrt{2}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x=2\sqrt{2}\end{align}

(ii)

Mean

\begin{align}\frac{\sqrt{32k^2}+\sqrt{50k^2}+\sqrt{128k^2}+\sqrt{98k^2}}{4}&=\frac{4\sqrt{2}k+8\sqrt{2}k+7\sqrt{2}k+5\sqrt{2}k}{4}\\&=\frac{24\sqrt{2}k}{4}\\&=6\sqrt{2}k\end{align}

$\,$

Median

\begin{align}\frac{1}{2}(\sqrt{50k^2}+\sqrt{50k^2})&=\frac{1}{2}(5\sqrt{2}k+7\sqrt{2}k)\\&=\frac{1}{2}(12\sqrt{2}k)\\&=6\sqrt{2}k\end{align}

Therefore, the mean and the median are the same.

Video Walkthrough

## 2019 Paper 2 Question 8

(a) A motoring magazine collected data on cars on a particular stretch of road.
Certain details on $$800$$ cars were recorded.

(i) The ages of the $$800$$ cars were recorded. $$174$$ of them were new (less than 1 year old).
Find the $$95\%$$ confidence interval for the proportion of new cars on this road.
Give your answer correct to $$4$$ significant figures.

(ii) The data on the speeds of these $$800$$ vehicles is normally distributed with an average speed of $$87.3$$ kilometres per hour and a standard deviation of $$12$$ kilometres per hour.
What proportion of cars on this stretch of road would you expect to find travelling at over $$95$$ kilometres per hour?

(iii) The driver of a car was told that $$70\%$$ of all the speeds recorded were higher than his
speed. Find the speed at which this driver was recorded.

(i) $$0.1889<p<0.2461$$

(ii) $$0.2611$$

(iii) $$81\mbox{ km/h}$$

Solution

(i)

\begin{align}\frac{174}{800}-1.96\sqrt{\frac{\frac{174}{800}\times\frac{626}{800}}{800}}<p<\frac{174}{800}+1.96\sqrt{\frac{\frac{174}{800}\times\frac{626}{800}}{800}}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}0.1889<p<0.2461\end{align}

(ii)

\begin{align}z&=\frac{x-\mu}{\sigma}\\&=\frac{95-87.3}{12}\\&=0.64166….\end{align}

\begin{align}\downarrow\end{align}

\begin{align}P(z\leq 0.64166…)=0.7389\end{align}

\begin{align}\downarrow\end{align}

\begin{align}P(z\geq 0.64166…)&=1-0.7389\\&=0.2611\end{align}

(iii)

\begin{align}-0.52=\frac{x-87.3}{12}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x&=(-0.52)(12)+87.3\\&\approx81\mbox{ km/h}\end{align}

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(b)

(i) A road safety programme was carried out in the area using posters, signs and radio slots. After the programme the motoring magazine recorded the speeds of $$100$$ passing cars. The magazine carried out a hypothesis test, at the $$5\%$$ level of significance, to determine whether the average speed had changed.
The $$p$$-value of the test was $$0.024$$.
What can the magazine conclude based on this $$p$$-value?

(ii) The magazine found that the average speed of this sample was lower than the previously established average speed of $$87.3$$ kilometres per hour.
Find the average speed of the cars in this sample, correct to $$1$$ decimal place.

(i) We can conclude that the average speed had changed as the $$p$$-value is less than $$0.05$$.

(ii) $$84.6\mbox{ km}$$

Solution

(i) We can conclude that the average speed had changed as the $$p$$-value is less than $$0.05$$.

(ii)

\begin{align}0.024=2(1-P)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}P=0.988\end{align}

\begin{align}\downarrow\end{align}

\begin{align}z=\pm2.26\end{align}

\begin{align}\downarrow\end{align}

\begin{align}z=-2.26\end{align}

as we are told that the mean is lower than the previous mean.

\begin{align}-2.26=\frac{x-87.3}{12/\sqrt{100}}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x&=87.3-\frac{2.26(12)}{10}\\&\approx84.6\mbox{ km}\end{align}

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## 2018 Paper 2 Question 2

(a) The diagram shows the standard normal curve. The shaded area represents $$67\%$$ of the data.
Find the value of $$z_1$$.

$$z_1=0.44$$

Solution

\begin{align}P(z<z_1)=0.67\end{align}

\begin{align}\downarrow\end{align}

\begin{align}z_1=0.44\end{align}

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(b) The percentage results in a Maths exam for a class had a mean mark of $$70$$ with a standard deviation of $$15$$. The percentage results in an English exam for the same class had a mean mark of $$72$$ with a standard deviation of $$10$$. The results in both exams were normally distributed.

(i) Mary got $$65$$ in Maths and $$68$$ in English. In which exam did Mary do better relative to the other students in the class? Justify your answer.

(ii) In English the top $$15\%$$ of students were awarded an A grade.
Find the least whole number mark that merited the award of an A grade in English.

(iii) Using the empirical rule, or otherwise, estimate the percentage of students in the class who scored between $$52$$ and $$82$$ in the English test.

(i) Since her $$z$$-score was larger for maths, Mary did better in maths.

(ii) $$83\%$$

(iii) $$81.5\%$$

Solution

(i)

Maths

\begin{align}z&=\frac{x-\mu}{\sigma}\\&=\frac{65-70}{15}=-\frac{1}{3}\end{align}

$\,$

English

\begin{align}z&=\frac{x-\mu}{\sigma}\\&=\frac{68-72}{10}=-\frac{2}{5}\end{align}

Since her $$z$$-score was larger for maths, Mary did better in maths.

(ii)

\begin{align}P(z>z_1)=0.15\end{align}

\begin{align}\downarrow\end{align}

\begin{align}z=1.04\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{x-72}{10}=1.04\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x&=72+(10)(1.04)\\&=82.4\%\end{align}

Therefore, the least whole number score is $$83\%$$.

(iii) As $$52$$ is $$\dfrac{95}{2}=47.5\%$$ below the mean, and as $$82$$ is $$\dfrac{68}{2}=34\%$$ above the mean, the percentage of students is

\begin{align}47.5+34=81.5\%\end{align}

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## 2018 Paper 2 Question 8(a) & (b)

Acme Confectionery makes cakes and chocolate bars.

(a)

(i) Acme Confectionery has launched a new bar called Chocolate Crunch. The weights of these new bars are normally distributed with a mean of $$4.64\mbox{ g}$$ and a standard deviation
of $$0.12\mbox{ g}$$. A sample of $$10$$ bars is selected at random and the mean weight of the sample is found.
Find the probability that the mean weight of the sample is between $$4.6\mbox{ g}$$ and $$4.7\mbox{ g}$$.

(ii) A company surveyed $$400$$ people, chosen from the population of people who had bought at least one Chocolate Crunch bar.
Of those surveyed, $$324$$ of them said they liked the new bar.
Create the $$95\%$$ confidence interval for the population proportion who liked the new bar.
Give your answer correct to $$2$$ decimal places.

(i) $$0.796$$

(ii) $$0.77\leq p\leq0.85$$

Solution

(i)

\begin{align}z_1&=\frac{x_1-\mu}{\sigma/\sqrt{n}}\\&=\frac{4.6-4.64}{0.12/\sqrt{10}}\\&=-1.05…\end{align}

and

\begin{align}z_2&=\frac{x_2-\mu}{\sigma/\sqrt{n}}\\&=\frac{4.7-4.64}{0.12/\sqrt{10}}\\&=1.58…\end{align}

\begin{align}\downarrow\end{align}

\begin{align}p(-1.05<z<1.58)&=0.9429-(1-0.8531)\\&=0.796\end{align}

(ii)

\begin{align}1.96\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}&=1.96\sqrt{\frac{(0.81)(1-0.81)}{400}}\\&=0.0384…\end{align}

$\,$

Confidence Interval

\begin{align}\hat{p}-0.0384…\leq p\leq\hat{p}+0.0384&=0.81-0.0384…\leq p\leq0.81+0.0384\\&\approx0.77\leq p\leq0.85\end{align}

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(b)

(i) State whether each of the following statements are Always True, Sometimes True or Never True.
In the statements, $$n$$ is the size of the sample and $$\hat{p}$$ is the sample proportion.

1. When forming confidence intervals (for fixed $$n$$ and $$\hat{p}$$), an increased confidence level implies a wider interval.
2. As the value of $$\hat{p}$$ increases (for fixed $$n$$), the estimated standard error of the population proportion increases.
3. As the value of $$\hat{p}(1-\hat{p})$$ increases (for fixed $$n$$), the estimated standard error of the population proportion increases.
4. As $$n$$, the number of people sampled, increases (for fixed $$\hat{p}$$, the estimated standard error of the population proportion increases.

(ii) Using calculus or otherwise, find the maximum value of $$\hat{p}(1-\hat{p})$$.

(iii) Hence, find the largest possible value of the radius of the $$95\%$$ confidence interval for a population proportion, given a random sample of size $$800$$.

(i)

1. Always true.
2. Sometimes true.
3. Always true
4. Never true

(ii) $$\dfrac{1}{4}$$

(ii) $$0.0346…$$

Solution

(i)

1. Always true.
2. Sometimes true.
3. Always true
4. Never true

(ii)

\begin{align}A(\hat{p})&=\hat{p}(1-\hat{p})\\&=\hat{p}-\hat{p}^2\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{dA}{d\hat{p}}&=1-2\hat{p}\end{align}

$\,$

Maximum

\begin{align}1-2\hat{p}=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\hat{p}=\frac{1}{2}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}A\left(\frac{1}{2}\right)&=\frac{1}{2}-\left(\frac{1}{2}\right)^2\\&=\frac{1}{4}\end{align}

(ii)

\begin{align}r&=1.96\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\\&=1.96\sqrt{\frac{\frac{1}{4}\left(1-\frac{1}{4}\right)}{800}}\\&=0.0346…\end{align}

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## 2017 Paper 2 Question 2(a) - (c)

An experiment measures the fuel consumption at various speeds for a particular model of car.
The data collected are shown in Table $$1$$ below.

Speed (km/hour) $$40$$ $$48$$ $$56$$ $$64$$ $$88$$ $$96$$ $$112$$

Fuel consumption (km/litre)

$$21$$

$$16$$

$$18$$

$$16$$

$$13$$

$$11$$

$$9$$

(a) Find the correlation coefficient of the data in Table $$1$$, correct to $$3$$ decimal places.

$$r=-0.957$$

Solution

$$r=-0.957$$

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(b) Plot the points from the table on the grid below and draw the line of best fit (by eye).

Solution
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(c) The slope of the line of best fit is found to be $$-0.15$$.
What does this value represent in the context of the data?

$$0.15$$ is the rate at which fuel consumption is decreasing as the speed increases.

Solution

$$0.15$$ is the rate at which fuel consumption is decreasing as the speed increases.

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## 2017 Paper 2 Question 8(a)

(a) In $$2015$$, in a particular country, the weights of $$15$$ year olds were normally distributed with a mean of $$63.5\mbox{ kg}$$ and a standard deviation of $$10\mbox{ kg}$$.

(i) In $$2015$$, Mariska was a $$15$$ year old in that country. Her weight was $$50\mbox{ kg}$$.
Find the percentage of $$15$$ year olds in that country who weighed more than Mariska.

(ii) In $$2015$$, Kamal was a $$15$$ year old in that country.
$$1.5\%$$ of $$15$$ year olds in that country were heavier than Kamal.
Find Kamal’s weight.

(iii) In $$2016$$, $$150$$ of the $$15$$ year olds in that country were randomly selected and their weights recorded. It was found that their weights were normally distributed with a mean weight of $$62\mbox{ kg}$$ and a standard deviation of $$10\mbox{ kg}$$. Test the hypothesis, at the $$5\%$$ level of significance, that the mean weight of $$15$$ year olds, in that country, had not changed from $$2015$$ to $$2016$$. State the null hypothesis and your alternative hypothesis.
Give your conclusion in the context of the question.

(i) $$0.9115$$

(ii) $$85.2\mbox{ kg}$$

(iii) The mean weight has not changed.

Solution

(i)

\begin{align}z&=\frac{x-\mu}{\sigma}\\&=\frac{50-63.5}{10}\\&=-1.35\end{align}

\begin{align}\downarrow\end{align}

\begin{align}P(z<1.35)=0.9115\end{align}

(ii)

\begin{align}P(x<z)&=1-0.015\\&=0.985\end{align}

\begin{align}\downarrow\end{align}

\begin{align}z=2.17\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{x-\mu}{\sigma}=2.17\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{x-63.5}{10}=2.17\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x&=(10)(2.17)+63.5\\&=85.2\mbox{ kg}\end{align}

(iii)

Null hypothesis: Mean weight had not changed.

Alternative hypothesis: Mean weight had changed.

Calculations:

\begin{align}z&=\frac{\bar{x}-\mu}{\sigma/\sqrt{n}}\\&=\frac{62-63.5}{10/\sqrt{150}}\\&=-1.837…\end{align}

Conclusion:

As $$-1.837…>-1.96$$, the mean weight has not changed.

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## 2016 Paper 2 Question 9

Data on earnings were published for a particular country. The data showed that the annual income of people in full-time employment was normally distributed with a mean of €$$39{,}400$$ and a standard deviation of €$$12{,}920$$.

(a)

(i) The government intends to impose a new tax on incomes over €$$60{,}000$$.
Find the percentage of full-time workers who will be liable for this tax, correct to one decimal place.

(ii) The government will also provide a subsidy to the lowest $$10\%$$ of income earners.
Find the level of income at which the government will stop paying the subsidy, correct to the nearest euro.

(iii) Some time later a research institute surveyed a sample of $$1000$$ full-time workers, randomly selected, and found that the mean annual income of the sample was €$$38{,}280$$.
Test the hypothesis, at the $$5\%$$ level of significance, that the mean annual income of full-time workers has changed since the national data were published.
State the null hypothesis and the alternative hypothesis.
Give your conclusion in the context of the question.

(i) $$5.6\%$$

(ii) $$22{,}862\mbox{ euro}$$

(iii) The mean income has changed.

Solution

(i)

\begin{align}z&=\frac{x-\mu}{\sigma}\\&=\frac{60{,}000-39{,}400}{12{,}920}\\&\approx1.59\end{align}

\begin{align}\downarrow\end{align}

\begin{align}P(z>1.59)&=1-P(z<1.59)\\&=1-0.9441\\&=0.0559\\&\approx5.6\%\end{align}

(ii)

\begin{align}z=-1.28\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{x-39{,}400}{12{,}920}=-1.28\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x&=39{,}400-1.28(12{,}920)\\&\approx22{,}862\mbox{ euro}\end{align}

(iii)

Null hypothesis: The mean income has not changed.

Alternative hypothesis: The mean income has changed

Calculations:

\begin{align}z&=\frac{\bar{x}-\mu}{\sigma/\sqrt{n}}\\&=\frac{38{,}280-39{,}400}{12{,}920/\sqrt{1{,}000}}\\&\approx-2.74\end{align}

Conclusion: As $$-2.74<-1.96$$, the mean income has changed.

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(b) The research institute surveyed $$400$$ full-time farmers, randomly selected from all the full-time farmers in the country, and found that the mean income for the sample was €$$26{,}974$$ and the standard deviation was €$$5120$$.
Assuming that annual farm income is normally distributed in this country, create a $$95\%$$ confidence interval for the mean income of full-time farmers.

$$26{,}472.24\leq\mu\leq27{,}475.76$$

Solution

\begin{align}\bar{x}-1.96\left(\frac{\sigma}{\sqrt{n}}\right)\leq\mu\leq\bar{x}+1.96\left(\frac{\sigma}{\sqrt{n}}\right)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}26{,}974-1.96\left(\frac{5{,}120}{\sqrt{400}}\right)\leq\mu\leq26{,}974+1.96\left(\frac{5{,}120}{\sqrt{400}}\right)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}26{,}472.24\leq\mu\leq27{,}475.76\end{align}

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(c) It is known that data on farm size are not normally distributed.
The research institute could take many large random samples of farm size and create a sampling distribution of the means of all these samples.
Give one reason why they might do this.

The sampling distribution will be a normal distribution.

Solution

The sampling distribution will be a normal distribution.

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(d) The research institute also carried out a survey into the use of agricultural land.
$$n$$ farmers were surveyed.
If the margin of error of the survey was $$4.5\%$$, find the value of $$n$$.

$$494$$

Solution

\begin{align}\frac{1}{\sqrt{n}}=0.045\end{align}

\begin{align}\downarrow\end{align}

\begin{align}n&=\frac{1}{0.045^2}\\&\approx494\end{align}

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## 2015 Paper 2 Question 2

A survey of $$100$$ shoppers, randomly selected from a large number of Saturday supermarket shoppers, showed that the mean shopping spend was €$$90.45$$. The standard deviation of this sample was €$$20.73$$.

(a) Find a $$95\%$$ confidence interval for the mean amount spent in a supermarket on that Saturday.

$$86.39\mbox{ euro}<\mu<94.51\mbox{ euro}$$

Solution

\begin{align}\bar{x}-1.96\frac{\sigma}{\sqrt{n}}<\mu<\bar{x}+1.96\frac{\sigma}{\sqrt{n}}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}90.45-1.96\frac{20.73}{\sqrt{100}}<\mu<90.45+1.96\frac{20.73}{\sqrt{100}}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}86.39\mbox{ euro}<\mu<94.51\mbox{ euro}\end{align}

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(b) A supermarket has claimed that the mean amount spent by shoppers on a Saturday is €$$94$$.
Based on the survey, test the supermarket’s claim using a $$5\%$$ level of significance. Clearly state your null hypothesis, your alternative hypothesis, and your conclusion.

The meant amount spent is $$94\mbox{ euro}$$.

Solution

Null hypothesis: Meant amount spent is $$94\mbox{ euro}$$.

Alternative hypothesis: Meant amount spent is not $$94\mbox{ euro}$$.

Calculations:

\begin{align}z&=\frac{\bar{x}-\mu}{\sigma}\\&=\frac{90.45-94}{2.073}\\&\approx-1.71\end{align}

Conclusion: As $$-1.71>-1.96$$, the meant amount spent is $$94\mbox{ euro}$$.

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(c) Find the $$p$$-value of the test you performed in part (b) above and explain what this value represents in the context of the question.

$$p\mbox{-value}=0.0872$$.

If the mean amount spent was €94, then the probability that the sample mean would be €$$90.45$$ by chance is $$8.72\%$$.

Solution

\begin{align}p\mbox{-value}&=2(1-P(z<1.71)\\&=2(1-0.9564)\\&=0.0872\end{align}

If the mean amount spent was €94, then the probability that the sample mean would be €$$90.45$$ by chance is $$8.72\%$$.

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