ZIMSEC A Level Statistics Paper 4 November 2009

Section a

Section a, Question 2

[2 marks]Expectation of a linear function

A random variable X has E(X) = 10 and Var(X) = 9. Find E(Y) where Y = 2X - 3.

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[2 marks]Variance of a linear function

A random variable X has E(X) = 10 and Var(X) = 9. Find Var(Y) where Y = 2X - 3.

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[3 marks]Geometric probability

The owners of a motel have noticed that in the long run 40% of the people who stop and inquire about a room for the night actually book a room. Inquiries are independent of one another.

How many inquiries must the owners answer to be 99% sure of at least one booking?

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[2 marks]Geometric probability

At a motel, 40% of the people who inquire about a room actually book one, and inquiries are independent of one another.

Find the probability that the owners get at least one booking from 4 inquiries.

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[3 marks]Selections without replacement

Three tickets for a musical show are sent to a high school musical club. Fifteen girls and ten boys would like a ticket, and the three people to receive a ticket are chosen at random from the 25.

Find the probability that the three chosen are exactly 2 boys and 1 girl.

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[3 marks]Selections without replacement

Three tickets for a musical show are sent to a high school musical club. Fifteen girls and ten boys would like a ticket, and the three people to receive a ticket are chosen at random from the 25.

Find the probability that at least 2 of the three chosen are girls.

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[2 marks]Stem and leaf diagrams

Which of these is an advantage of using a stem and leaf diagram rather than a grouped frequency table to analyse a set of data?

AIt replaces the individual values by their class mid-points, which makes the calculation of the mean and the standard deviation considerably shorter.
BIt plots the values against time, which shows any trend or seasonal pattern in the data that a grouped frequency table would conceal.
CIt keeps every original data value, so the median, the quartiles and the range can all be read straight off it while the shape of the distribution is still visible.
DIt groups the values into classes of equal width, which makes the total frequency easier to check than counting the raw values one at a time.

[1 marks]Measures of central tendency

In which situation is the mode the appropriate measure of central tendency?

AWhen the data are numerical but contain a few extreme outliers, such as the salaries in a firm where one director earns far more than anyone else.
BWhen the data are categorical, so that adding and ordering the values make no sense, such as the colours of the shirts sold in a shop.
CWhen the data are numerical and every value in the set happens to occur exactly the same number of times, such as a complete set of examination scores.
DWhen the data are continuous measurements that are roughly symmetric about their centre, such as the heights of a large group of adults.

[3 marks]Normal distribution, percentiles

The masses of a group of adult males are normally distributed with mean 65 kg and standard deviation 10 kg. Males are considered overweight if they are in the top 5% of the group by mass.

Find the least mass, in kg, to be considered overweight.

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[2 marks]Chi-squared test of association

A random sample of 400 students was asked their view on infusion of environmental issues in their college curriculum.

	In favour	Opposed	Undecided
Females	115	60	36
Males	90	85	14

A chi-squared test is to be carried out. Calculate the expected frequency of females who are Undecided.

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[3 marks]Chi-squared test of association

A random sample of 400 students was asked their view on infusion of environmental issues in their college curriculum.

	In favour	Opposed	Undecided
Females	115	60	36
Males	90	85	14

Calculate the value of the chi-squared test statistic for testing whether there is any difference in opinion between males and females.

Answer this when you sit the paper.

[2 marks]Chi-squared test of association

A chi-squared test at the 5 % level is carried out on a 2 by 3 table classifying 400 students by sex and by opinion. The calculated test statistic is 15.88.

What are the degrees of freedom and the correct conclusion?

A2 degrees of freedom, critical value 5.991; since 15.88 is greater, accept the null hypothesis: there is no difference in opinion between males and females.
B5 degrees of freedom, critical value 11.07; since 15.88 is greater, reject the null hypothesis: there is a difference in opinion between males and females.
C1 degree of freedom, critical value 3.841; since 15.88 is greater, reject the null hypothesis: being male is what causes a student to be opposed.
D2 degrees of freedom, critical value 5.991; since 15.88 is greater, reject the null hypothesis: there is a difference in opinion between males and females.

Section a, Question 3

[2 marks]Continuous random variables

The duration X minutes of a telephone call is a continuous random variable with probability density function $f(x)=x^{-2}$ for $x \ge 1$ and $f(x)=0$ otherwise.

Which of these is the cumulative distribution function F(x) for $x \ge 1$ ?

A $F(x)=\dfrac{1}{x}-1$
B $F(x)=\dfrac{1}{x}$
C $F(x)=1-\dfrac{1}{x}$
D $F(x)=1-\dfrac{1}{x^{2}}$

[2 marks]Continuous random variables

The duration X minutes of a telephone call is a continuous random variable with probability density function $f(x)=x^{-2}$ for $x \ge 1$ and $f(x)=0$ otherwise.

Find P(X > 5), giving your answer as an exact fraction.

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[3 marks]Conditional probability with a continuous variable

The duration X minutes of a telephone call is a continuous random variable with probability density function $f(x)=x^{-2}$ for $x \ge 1$ and $f(x)=0$ otherwise.

Given that a call has already lasted for 5 minutes, find the conditional probability that its total duration will be less than 7 minutes.

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[2 marks]Median

The bus fares, in thousands of dollars, paid by 19 football fans selected at random from a football crowd were

73, 85, 48, 80, 53, 75, 55, 58, 62, 69, 63, 64, 73, 65, 55, 54, 55, 45, 55.

Find the median fare.

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[2 marks]Quartiles

The bus fares, in thousands of dollars, paid by 19 football fans selected at random from a football crowd were

73, 85, 48, 80, 53, 75, 55, 58, 62, 69, 63, 64, 73, 65, 55, 54, 55, 45, 55.

Find the lower quartile of the fares.

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[3 marks]Interquartile range

The bus fares, in thousands of dollars, paid by 19 football fans selected at random from a football crowd were

73, 85, 48, 80, 53, 75, 55, 58, 62, 69, 63, 64, 73, 65, 55, 54, 55, 45, 55.

Find the interquartile range of the fares.

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[2 marks]Hypothesis testing

A manufacturer claims that a new machine has improved his product. The old machine is known to have given 20% defectives. In a sample of 20 items from the new machine, 2 were found to be defective. A binomial test is carried out at the 5% significance level.

What are the appropriate hypotheses?

A $H_{0}: p = 0.20$ against $H_{1}: p \ne 0.20$ , a two-tailed test, since nothing in the claim says in which direction the defective rate is supposed to have moved.
B $H_{0}: p = 0.20$ against $H_{1}: p < 0.20$ , a one-tailed test, since an improved product means a smaller proportion of defectives than the old machine's 20 per cent.
C $H_{0}: p = 0.20$ against $H_{1}: p > 0.20$ , a one-tailed test, since the manufacturer is claiming that the defective rate has changed from its old value.
D $H_{0}: p = 0.10$ against $H_{1}: p < 0.10$ , a one-tailed test, since 2 defectives out of 20 is a rate of 10 per cent in the sample that was actually taken.

[3 marks]Binomial hypothesis test

Calculate the probability of 2 or fewer defectives in a sample of 20 if the defective rate is still 20%.

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[2 marks]Binomial hypothesis test

A binomial test of $H_{0}: p = 0.20$ against $H_{1}: p < 0.20$ is carried out at the 5 % level on a sample of 20 items in which 2 were defective. Under the null hypothesis $P(X \le 2) = 0.2061$ .

What is the correct conclusion?

ASince only 2 items out of 20 were defective, reject the null hypothesis: a sample rate of 10 per cent is plainly half the old machine's rate of 20 per cent.
BSince 0.2061 is greater than 0.05, do not reject the null hypothesis: there is insufficient evidence at the 5 % level to support the claim that the product has improved.
CSince 0.2061 is greater than 0.05, reject the null hypothesis: there is evidence at the 5 % level that the product has improved as the manufacturer claims.
DSince 0.2061 is greater than 0.05, do not reject the null hypothesis: this proves that the new machine gives exactly the same defective rate as the old one.

Section a, Question 4

[2 marks]Discrete random variables

A triangular prism has two equilateral triangular faces and three rectangular faces. The rectangular faces are numbered 1, 2 and 3 and the triangular faces are numbered 4 and 5. When the prism is tossed, the probability that it lands on each rectangular face is 2k and the probability that it lands on each triangular face is k.

Calculate the value of k.

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[3 marks]Discrete random variables

X is the number on which the prism lands. Calculate $E(X^{2})$ .

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[3 marks]Discrete random variables

X is the number on which the prism lands, and $E(X) = 2\dfrac{5}{8}$ . Find Var(X).

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Section a, Question 5

[2 marks]Regression summary statistics

Two judges, A and B, independently awarded marks x and y respectively to 10 architectural designs. The marks are summarised by $n=10$ , $\sum x = 600$ , $\sum x^{2} = 39\,900$ , $\sum y = 526$ , $\sum y^{2} = 29\,548$ and $\sum xy = 34\,145$ .

Calculate $S_{xy}$ .

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[3 marks]Product moment correlation coefficient

Calculate the product-moment correlation coefficient between the marks awarded by the two judges.

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[3 marks]Regression line of y on x

Find the gradient of the regression line of y on x.

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[2 marks]Unbiased estimate of a population mean

A random sample of size 40 is taken from a population of fish. X denotes the length of a fish in centimetres, and the sample is summarised by $\sum(x-20) = 19$ and $\sum(x-20)^{2} = 68$ .

Find the unbiased estimate of the population mean, in centimetres.

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[2 marks]Unbiased estimate of a population variance

A random sample of size 40 is taken from a population of fish. X denotes the length of a fish in centimetres, and the sample is summarised by $\sum(x-20) = 19$ and $\sum(x-20)^{2} = 68$ .

Find the unbiased estimate of the population variance.

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[2 marks]Confidence intervals

The 95% confidence interval for the mean life of a brand of light bulbs, constructed from a sample of size 36, is (1023.3 hrs; 1161.7 hrs).

Find the mean life of the bulbs in that sample, in hours.

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[3 marks]Confidence intervals

The 95% confidence interval for the mean life of a brand of light bulbs, constructed from a sample of size 36, is (1023.3 hrs; 1161.7 hrs). The life of the bulbs is normally distributed.

Find the upper limit of the 99% confidence interval for the mean life, in hours.

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Section a, Question 6

[2 marks]Normal approximation to the binomial

Data shows that 42% of Zimbabweans eat breakfast every day. A random sample of 300 Zimbabweans is taken and X is the number in the sample who eat breakfast every day.

Calculate the standard deviation of X.

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[2 marks]Normal approximation to the binomial

A binomial variable X with $n = 300$ and $p = 0.42$ is approximated by a normal distribution with mean 126 and standard deviation 8.549.

Which value should be standardised to find $P(X \le 100)$ ?

A100.5, because a discrete count is being approximated by a continuous distribution and the bar for 100 stretches to 100.5.
B99.5, because a discrete count is being approximated by a continuous distribution and the bar for 100 begins at 99.5.
C101, because the next whole number above the one named in the inequality is where the normal curve should be cut.
D100, because the inequality already includes the value 100 and so no adjustment to it is needed at all.

[3 marks]Normal approximation to the binomial

Data shows that 42% of Zimbabweans eat breakfast every day. A random sample of 300 Zimbabweans is taken and X is the number in the sample who eat breakfast every day.

Find the probability that at most 100 of the sample eat breakfast every day.

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[3 marks]Normal approximation to the binomial

Data shows that 42% of Zimbabweans eat breakfast every day. A random sample of 300 Zimbabweans is taken and X is the number in the sample who eat breakfast every day.

Find the probability that from 130 to 140 of the sample eat breakfast every day.

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[2 marks]Fitting a Poisson distribution

The number of computer system breakdowns per month at a University was observed over 100 months.

Breakdowns (X)	0	1	2	3	4	5 or more
Frequency	15	25	30	21	9	0

Calculate the mean number of breakdowns per month.

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[2 marks]Fitting a Poisson distribution

The number of computer system breakdowns per month at a University was observed over 100 months.

Breakdowns (X)	0	1	2	3	4	5 or more
Frequency	15	25	30	21	9	0

A Poisson distribution of mean 1.84 is fitted to these data. Calculate the expected number of months, out of 100, with exactly 2 breakdowns.

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[2 marks]Chi-squared goodness of fit

A Poisson distribution of mean 1.84 is fitted to 100 months of breakdown data. The expected frequencies are 15.88, 29.22, 26.88, 16.49 and 7.59 for 0, 1, 2, 3 and 4 breakdowns, and 3.94 for 5 or more.

The classes with an expected frequency below 5 must be pooled with their neighbour. Calculate the expected frequency of the pooled class "4 or more".

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[3 marks]Chi-squared goodness of fit

For 100 months of breakdown data the observed frequencies are 15, 25, 30, 21 and 9 for 0, 1, 2, 3 and 4 or more breakdowns, and the expected frequencies under a Poisson model of mean 1.84 are 15.88, 29.22, 26.88, 16.49 and 11.52.

Calculate the value of the chi-squared goodness of fit statistic.

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[3 marks]Chi-squared goodness of fit

A Poisson goodness of fit test is carried out at the 5 % level on 100 months of breakdown data. After pooling there are 5 classes, the mean was estimated from the data, and the calculated chi-squared statistic is 2.81.

What are the degrees of freedom and the correct conclusion?

A3 degrees of freedom, critical value 7.815; since 2.81 is smaller, do not reject the null hypothesis, so the breakdowns are consistent with a Poisson distribution.
B5 degrees of freedom, critical value 11.07; since 2.81 is smaller, the breakdowns are proved beyond doubt to follow a Poisson distribution exactly.
C4 degrees of freedom, critical value 9.488; since 2.81 is smaller, do not reject the null hypothesis, so the breakdowns are consistent with a Poisson distribution.
D3 degrees of freedom, critical value 7.815; since 2.81 is smaller, reject the null hypothesis, so the breakdowns do not follow a Poisson distribution.

Statistics Paper 4 November 2009

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The questions

Section a

Section a, Question 2

Section a, Question 3

Section a, Question 4

Section a, Question 5

Section a, Question 6

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The answers, and why they are the answers

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