ZIMSEC A Level Statistics Paper 4 June 2011

Section a

Section a, Question 1

[2 marks]Geometric distribution

A discrete random variable Y follows a geometric distribution with variance 12. Find the probability of success on a single trial.

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[2 marks]Geometric distribution

A discrete random variable Y follows a geometric distribution with probability of success

\dfrac{1}{4}

on each trial. Find the probability that Y exceeds 3.

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Section a, Question 2

[2 marks]Conditional probability

Of 50 patients at a clinic, 15 are selected at random for a new dietary treatment and the other 35 get the standard drug treatment. The probability of a cure is 0.65 with the standard drug and 0.95 with the new treatment. One of the treated patients is then selected at random. Find the probability that the patient was cured.

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[3 marks]Conditional probability

Of 50 patients at a clinic, 15 are selected at random for a new dietary treatment and the other 35 get the standard drug treatment. The probability of a cure is 0.65 with the standard drug and 0.95 with the new treatment. One of the treated patients is selected at random and found to have been cured. Find the probability that she received the new dietary treatment.

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Section a, Question 3

[2 marks]Discrete random variables

Two unbiased tetrahedral dice with faces numbered 1, 2, 3, 4 are thrown, and the score X is the sum of the two numbers shown. Find

P(X=5)

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[2 marks]Discrete random variables

Two unbiased tetrahedral dice with faces numbered 1, 2, 3, 4 are thrown, and the score X is the sum of the two numbers shown. Calculate

E(X)

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[2 marks]Discrete random variables

Two unbiased tetrahedral dice with faces numbered 1, 2, 3, 4 are thrown, and the score X is the sum of the two numbers shown. Calculate

\operatorname{Var}(X)

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Section a, Question 4

[3 marks]Continuous random variables

X is a continuous random variable which follows a rectangular distribution over the interval

[a,b]

where

a<b

. Given that

P(X>4)=0,5

and

P(X<5,6)=0,9

, find the values of

a

and

b

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[2 marks]Continuous random variables

X is a continuous random variable which follows a rectangular distribution over the interval

[2,6]

. Calculate

\operatorname{Var}(X)

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[1 marks]Continuous random variables

X follows a rectangular distribution over the interval

[a,b]

and

P(X>4)=0,5

. What does this tell you about

a

and

b

AThat 4 is the midpoint of the interval, so $a+b=8$ .
BThat $a=4$ , since half the distribution lies above the lower limit.
CThat $b=8$ , since the distribution is symmetrical about the upper limit.
DThat $b-a=4$ , so the interval has length 4.

Section a, Question 5

[3 marks]Sampling distributions

Samples of size

n

are taken from a population which follows a normal distribution with mean 70 and standard deviation 5. Find

n

if the probability that the sample mean exceeds 68 is 0.9254.

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[1 marks]Confidence intervals

A random sample of 100 adults drank a mean of 7 100 ml of water in one week, with a standard deviation of 400 ml. Calculate the standard error of the sample mean, in ml.

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[3 marks]Confidence intervals

A random sample of 100 adults in Muzarabani drank a mean of 7 100 ml of water in one week, with a standard deviation of 400 ml. Calculate the 95% confidence interval in which the mean weekly consumption of water will lie. Give both limits in ml correct to 1 decimal place.

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Section a, Question 6

[3 marks]Hypothesis testing

On average 45% of those taking a driving test in Chegutu pass. In one particular week 100 examinees took the test and 40 passed. Taking

H_0: p=0,45

, calculate the value of the test statistic

z

, correct to 2 decimal places.

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[2 marks]Hypothesis testing

A claim that examiners were too harsh in one week is tested as

H_0: p=0,45

against

H_1: p<0,45

at the 5% level of significance, using a normal test statistic. State the critical value of

z

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[2 marks]Hypothesis testing

On average 45% of driving tests in Chegutu are passed. In one week 40 out of 100 examinees passed, and an examinee complained that the examiners were too harsh that week. A test of

H_0: p=0,45

against

H_1: p<0,45

at the 5% level gives

z=-1,01

against a critical value of

-1,645

. What is the conclusion?

AThe test is invalid, because a single week is not a random sample.
BReject $H_0$ : the pass rate that week was significantly above 45%.
CDo not reject $H_0$ : there is no evidence the examiners were harsher than usual.
DReject $H_0$ : the pass rate that week was significantly below the usual 45%, so the examinee's complaint stands.

Section a, Question 7

[2 marks]Poisson distribution

A school has 1 500 students who come to school every day, and the probability that a student is late on a particular day is 0.002. The number of late students in a day is modelled by a Poisson distribution. State the mean of that distribution.

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[2 marks]Poisson distribution

The number of students late at a school in a day follows a Poisson distribution with mean 3. Find, correct to 3 decimal places, the probability that on any given day at least one student will be late.

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[2 marks]Poisson distribution

The number of students late at a school in a day follows a Poisson distribution with mean 3, independently from day to day. Find, correct to 3 decimal places, the probability that no student will be late on two consecutive days.

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[2 marks]Poisson distribution

The number of students late at a school in a day follows a Poisson distribution with mean 3, independently from day to day. Find, correct to 3 decimal places, the probability that in a 5 day week there will be exactly 16 cases of late coming.

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Section a, Question 8

[2 marks]Chi-squared test of association

In a poll of 300 voters, party B received 105 votes in total and the Central region cast 95 votes in total. In a chi-squared test of association between party and region, calculate the expected number of Central-region votes for party B if there were no association.

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[2 marks]Chi-squared test of association

A chi-squared test of association is carried out on a contingency table of 3 parties by 3 regions. State the number of degrees of freedom.

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[3 marks]Chi-squared test of association

In a poll of 300 voters, the observed counts by party (rows A, B, C) and region (columns North, South, Central) were A: 35, 40, 25; B: 33, 30, 42; C: 37, 30, 28. The corresponding expected counts under no association are A: 35, 33,33, 31,67; B: 36,75, 35, 33,25; C: 33,25, 31,67, 30,08. Calculate the value of

\chi^{2}=\sum\dfrac{(O-E)^{2}}{E}

, correct to 2 decimal places.

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[2 marks]Chi-squared test of association

A chi-squared test of association between the party a voter chose and the region the voter lives in gives

\chi^{2}=6,79

with 4 degrees of freedom, against a 5% critical value of

9,488

. What is the conclusion at the 5% level of significance?

AReject the null hypothesis: there is clear evidence of an association between the party a voter chose and the region.
BReject the null hypothesis: the party a voter chose is independent of the region.
CThe test cannot be carried out, because some observed counts are below 40.
DDo not reject the null hypothesis: there is no evidence of an association.

Section a, Question 9

[3 marks]Normal distribution

The masses of professional soccer players are normally distributed with a mean of 66 kg and standard deviation

\sigma

. Given that 10% of the players have masses which exceed 72 kg, find the value of

\sigma

correct to 3 significant figures.

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[2 marks]Normal distribution

The masses of professional soccer players are normally distributed with mean 66 kg and standard deviation 4,68 kg. Find the probability that the mass of a randomly chosen player is at most 63 kg, correct to 4 decimal places.

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[1 marks]Normal distribution

The probability that a professional soccer player weighs at most 63 kg is 0,2608. Eleven players are selected at random, and Y is the number of them weighing at most 63 kg. What is the distribution of Y?

A $Y\sim\text{Po}(11\times0,2608)$
B $Y\sim B(11;\ 0,2608)$
C $Y\sim B(0,2608;\ 11)$
D $Y\sim N(11\times0,2608;\ 11\times0,2608\times0,7392)$

[3 marks]Normal distribution

The probability that a professional soccer player weighs at most 63 kg is 0,2608. Eleven players are randomly selected for a match. Find the probability that at least 3 of them weigh at most 63 kg, correct to 3 decimal places.

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Section a, Question 10

[2 marks]Continuous random variables

A continuous random variable X has probability density function

f(x)=k

for

0\le x\le3

f(x)=k(4-x)

for

3<x\le4

, and

f(x)=0

otherwise. Find the value of

k

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[3 marks]Continuous random variables

A continuous random variable X has probability density function

f(x)=\dfrac{2}{7}

for

0\le x\le3

f(x)=\dfrac{2}{7}(4-x)

for

3<x\le4

, and

f(x)=0

otherwise. Find

E(X)

, correct to 2 decimal places.

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[2 marks]Continuous random variables

A continuous random variable X has probability density function

f(x)=\dfrac{2}{7}

for

0\le x\le3

f(x)=\dfrac{2}{7}(4-x)

for

3<x\le4

, and

f(x)=0

otherwise. Find the median value of X.

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[1 marks]Continuous random variables

A continuous random variable X has probability density function

f(x)=\dfrac{2}{7}

for

0\le x\le3

f(x)=\dfrac{2}{7}(4-x)

for

3<x\le4

, and

f(x)=0

otherwise. Find

E(X^{2})

, correct to 2 decimal places.

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[3 marks]Continuous random variables

A continuous random variable X has

E(X)=\dfrac{37}{21}

and

E(X^{2})=\dfrac{25}{6}

. Find the standard deviation of X, correct to 2 decimal places.

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Section a, Question 11

[3 marks]Normal distribution

The mass of a toffee sweet has a normal distribution with mean 3.9 g and standard deviation 0.11 g. Find the probability that a randomly chosen toffee sweet weighs more than 4 g, correct to 3 decimal places.

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[2 marks]Normal distribution

The probability that a toffee sweet weighs more than 4 g is 0,182. Find the probability that of two randomly chosen toffee sweets, one weighs more than 4 g and the other weighs less than 4 g. Give your answer correct to 3 decimal places.

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[3 marks]Linear combinations of random variables

The mass of a toffee sweet has a normal distribution with mean 3.9 g and standard deviation 0.11 g. Find the probability that five randomly chosen toffee sweets weigh a total of more than 20 g, correct to 3 decimal places.

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[3 marks]Linear combinations of random variables

The mass of a toffee sweet is normally distributed with mean 3.9 g and standard deviation 0.11 g, and the mass of a mint is normally distributed with mean 5 g and standard deviation 0.16 g. Find the probability that the total mass of five randomly chosen toffee sweets is more than the total mass of four randomly chosen mints, correct to 3 decimal places.

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Section a, Question 12

[1 marks]Regression and correlation

For ten students,

x

is the Maths mark and

y

the Shona mark, with

n=10

\sum x=52

\sum y=55

and

\sum xy=204

. Calculate

S_{xy}

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[3 marks]Regression and correlation

Ten students sat a Maths test and a Shona test. Their Maths marks

x

were 8, 10, 2, 7, 3, 4, 5, 4, 8, 1 and their Shona marks

y

were 3, 1, 10, 3, 7, 8, 7, 6, 1, 9 respectively. Find the equation of the regression line of

y

x

, giving the coefficients correct to 3 significant figures.

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[2 marks]Regression and correlation

The regression line of the Shona mark

y

on the Maths mark

x

for ten students is

y=10,99-1,057x

, and the Maths marks in the sample ran from 1 to 10. Use the line to estimate the Shona mark of a student who gets 6 in Maths.

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[2 marks]Regression and correlation

The regression line of the Shona mark

y

on the Maths mark

x

for ten students is

y=10,99-1,057x

. The Maths marks in the sample ran from 1 to 10, and each test was marked out of 10. Why can the line not be used to estimate the Shona mark of a student who got 0 in Maths?

ABecause a regression line of $y$ on $x$ can never be used to predict a value of $y$ .
BBecause the correlation is too strong for the line to be reliable at the ends of the range.
CBecause $x=0$ is outside the range of the data, and the line predicts 10,99, above the maximum mark.
DBecause the gradient of the fitted line is negative, and a negative gradient may only be used for values of $x$ above the mean.

[3 marks]Regression and correlation

Ten students sat a Maths test and a Shona test. Their Maths marks

x

were 8, 10, 2, 7, 3, 4, 5, 4, 8, 1 and their Shona marks

y

were 3, 1, 10, 3, 7, 8, 7, 6, 1, 9 respectively. Find the product moment correlation coefficient, correct to 3 significant figures.

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[2 marks]Regression and correlation

For ten students, the product moment correlation coefficient between the Maths mark and the Shona mark is

r=-0,948

. What does this tell you about the conception that students who do well in Maths perform badly in Shona?

AIt supports it: the relationship is strongly negative, so high Maths marks go with low Shona marks.
BIt supports it, and it proves that being good at Maths is what causes a student to perform badly in Shona.
CIt says nothing either way, since a correlation coefficient cannot be negative.
DIt contradicts it: a value near $-1$ means the two marks are unrelated.

Statistics Paper 4 June 2011

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The questions

Section a

Section a, Question 1

Section a, Question 2

Section a, Question 3

Section a, Question 4

Section a, Question 5

Section a, Question 6

Section a, Question 7

Section a, Question 8

Section a, Question 9

Section a, Question 10

Section a, Question 11

Section a, Question 12

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The answers, and why they are the answers

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