ZIMSEC A Level Statistics Paper 4 November 2011

Section a

Section a, Question 1

[2 marks]Probability

A roulette wheel contains 38 numbers of which 18 are red, 18 are black and 2 are green. When the wheel is spun it is equally likely to land on any of the 38 numbers, and successive spins are independent. The wheel is spun twice. Find the probability that the ball lands on red both times, giving your answer correct to 4 decimal places.

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[2 marks]Probability

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Section a, Question 2

[2 marks]Continuous random variables

The continuous random variable X has probability density function

f(x)=kx

for

0\le x\le1

f(x)=k

for

1<x\le2

and

f(x)=0

otherwise, where

k

is a constant. Find the value of

k

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[3 marks]Continuous random variables

The continuous random variable X has probability density function

f(x)=kx

for

0\le x\le1

f(x)=k

for

1<x\le2

and

f(x)=0

otherwise, where

k

is a constant. Given that

k=\dfrac{2}{3}

, find the median

m

of X.

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Section a, Question 3

[3 marks]Binomial distribution

In a chemical industry workmen had a 20% chance of suffering from an occupational disease, independently of one another. Find the smallest number of workmen who could have been selected at random before the probability that at least one of them contracted the disease became greater than 0.9.

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[2 marks]Binomial distribution

In a chemical industry each workman has a 20% chance of suffering from an occupational disease, independently of the others. Ten workmen are selected at random. Find the probability that at least one of them contracted the disease, giving your answer correct to 4 decimal places.

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Section a, Question 4

[3 marks]Continuous random variables

After some rain the depth of moisture,

X

metres, in Arda Gardens is a continuous random variable with probability density function

f(x)=\dfrac{12x}{5}(b-x)

for

0\le x\le1

and

f(x)=0

otherwise. Find the value of

b

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[3 marks]Continuous random variables

After some rain the depth of moisture,

X

metres, in Arda Gardens is a continuous random variable with probability density function

f(x)=\dfrac{12x}{5}(b-x)

for

0\le x\le1

and

f(x)=0

otherwise. Given that

b=1.5

, calculate the probability that the depth of moisture exceeds 0.9, giving your answer correct to 4 decimal places.

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Section a, Question 5

[2 marks]Conditional probability

A simple weather model classifies each day as either fine or rainy. The probability that a fine day is followed by another fine day is 0.8, and the probability that a rainy day is followed by a fine day is 0.4. The probability that 1 February is fine is 0.75. Find the probability that 2 February is fine.

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[2 marks]Conditional probability

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[2 marks]Conditional probability

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Section a, Question 6

[1 marks]Normal approximation to the binomial distribution

Under which conditions may a normal distribution be used to approximate a binomial distribution

B(n,p)

AWhen $p=\dfrac{1}{2}$ , whatever the value of $n$ , since the distribution is then exactly symmetrical.
BWhen $n$ is large and $p$ is small, so that $np$ stays below about 5 and the mean and the variance are nearly equal.
CWhen $n$ is large and both $np>5$ and $n(1-p)>5$ .
DWhen $n$ is small, whatever the value of $p$ , since a short run of trials is always close to normal.

[2 marks]Normal approximation to the binomial distribution

It is estimated that 20% of people undergoing medical review are men. A random sample of 100 people undergoing a medical review is taken, and the number of men is approximated by a normal distribution. Find the standard deviation of that approximating normal distribution.

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[3 marks]Normal approximation to the binomial distribution

It is estimated that 20% of people undergoing medical review are men. A random sample of 100 people undergoing a medical review is taken, and the number of men is approximated by a normal distribution. Using a continuity correction, find the probability that more than 30 of the 100 are men.

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Section a, Question 7

[2 marks]Discrete random variables

An unbiased tetrahedral die has the number 1 on one face, the number 2 on another face and the number 3 on the remaining two faces, so a single throw scores 1 with probability

\frac14

, 2 with probability

\frac14

and 3 with probability

\frac12

. The die is thrown twice and X is the product of the two scores. Find

P(X=2)

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[1 marks]Discrete random variables

An unbiased tetrahedral die has the number 1 on one face, the number 2 on another face and the number 3 on the remaining two faces, so a single throw scores 1 with probability

\frac14

, 2 with probability

\frac14

and 3 with probability

\frac12

. The die is thrown twice and X is the product of the two scores. Find

P(X=9)

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[2 marks]Discrete random variables

An unbiased tetrahedral die has the number 1 on one face, the number 2 on another face and the number 3 on the remaining two faces, so a single throw scores 1 with probability

\frac14

, 2 with probability

\frac14

and 3 with probability

\frac12

. The die is thrown twice and X is the product of the two scores. Find

E(X)

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[3 marks]Discrete random variables

An unbiased tetrahedral die has the number 1 on one face, the number 2 on another face and the number 3 on the remaining two faces, so a single throw scores 1 with probability

\frac14

, 2 with probability

\frac14

and 3 with probability

\frac12

. The die is thrown twice and X is the product of the two scores. Find

\operatorname{Var}(X)

, giving your answer correct to 2 decimal places.

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Section a, Question 8

[2 marks]Poisson distribution

The number of patients admitted at a medical centre each day has a Poisson distribution with mean 2. Evaluate the probability that on a particular day there will be no admissions, giving your answer correct to 4 decimal places.

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[3 marks]Poisson distribution

The number of patients admitted at a medical centre each day has a Poisson distribution with mean 2. At the beginning of one day the hospital has five beds available. Calculate the probability that this will be an insufficient number for the day.

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[3 marks]Poisson distribution

The number of patients admitted at a medical centre each day has a Poisson distribution with mean 2. Calculate the probability that there will be exactly three admissions altogether on two consecutive days.

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[3 marks]Poisson distribution

At a medical centre 150 patients are attended to on a particular day and the probability that any one patient will be admitted is 0.02, independently of the others. Using a suitable approximation, find the probability that exactly 4 patients are admitted.

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[2 marks]Poisson distribution

150 patients are attended to at a medical centre on a particular day, and each is admitted with probability 0.02, independently of the others. Why is a Poisson distribution with mean 3 a suitable approximation to the exact binomial distribution here?

ABecause $n$ is large and $p$ is small, with $np=3$ moderate.
BBecause the success probability is close to $\dfrac{1}{2}$ , which makes the exact binomial distribution symmetrical.
CBecause $np(1-p)=2.94$ , and a variance below 5 is the condition for a Poisson approximation.
DBecause the number of trials is small, so the binomial coefficients would otherwise be awkward to work out by hand.

Section a, Question 9

[3 marks]Normal distribution

Boxes marked B contain big fruits and boxes marked S contain small fruits. The masses of the boxes are independent normal random variables: a big box has mean 10 kg and standard deviation 2 kg, and a small box has mean 12 kg and standard deviation 3 kg. Find the probability that the mass of a box marked S is less than 10 kg, giving your answer correct to 4 decimal places.

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[2 marks]Linear combinations of random variables

Boxes marked B contain big fruits and boxes marked S contain small fruits. The masses of the boxes are independent normal random variables: a big box has mean 10 kg and standard deviation 2 kg, and a small box has mean 12 kg and standard deviation 3 kg. Find the mean total mass, in kg, of 4 big fruit boxes and 5 small fruit boxes.

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[2 marks]Linear combinations of random variables

Boxes marked B contain big fruits and boxes marked S contain small fruits. The masses of the boxes are independent normal random variables: a big box has mean 10 kg and standard deviation 2 kg, and a small box has mean 12 kg and standard deviation 3 kg. Find the variance of the total mass of 4 big fruit boxes and 5 small fruit boxes.

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[3 marks]Linear combinations of random variables

Boxes marked B contain big fruits and boxes marked S contain small fruits. The masses of the boxes are independent normal random variables: a big box has mean 10 kg and standard deviation 2 kg, and a small box has mean 12 kg and standard deviation 3 kg. Find the probability that the total mass of 4 big fruit boxes and 5 small fruit boxes is greater than 90 kg, giving your answer correct to 3 decimal places.

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[3 marks]Normal distribution

B_{1}

and

B_{2}

are the masses of two independent boxes marked B. Find the value of

m

, in kg, such that

P\left(B_{1}+B_{2}<m\right)=\dfrac{1}{4}

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Section a, Question 10

[2 marks]Chi-squared goodness of fit test

A sports director compares form one interests in sporting disciplines with the form two interest distribution, which is Cricket 21.1%, Hockey 27.0%, Rugby 33.9% and Soccer 18.0%. A random sample of 200 form ones gave the frequencies Cricket 42, Hockey 62, Rugby 64 and Soccer 32. Find the expected frequency for Hockey under the hypothesis that the form one distribution matches the form two distribution.

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[2 marks]Chi-squared goodness of fit test

A sports director compares form one interests in sporting disciplines with the form two interest distribution, which is Cricket 21.1%, Hockey 27.0%, Rugby 33.9% and Soccer 18.0%. A random sample of 200 form ones gave the frequencies Cricket 42, Hockey 62, Rugby 64 and Soccer 32. Find the expected frequency for Rugby under the hypothesis that the form one distribution matches the form two distribution.

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[3 marks]Chi-squared goodness of fit test

A sports director compares form one interests in sporting disciplines with the form two interest distribution, which is Cricket 21.1%, Hockey 27.0%, Rugby 33.9% and Soccer 18.0%. A random sample of 200 form ones gave the frequencies Cricket 42, Hockey 62, Rugby 64 and Soccer 32. Calculate the value of the chi-squared test statistic, giving your answer correct to 2 decimal places.

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[1 marks]Chi-squared goodness of fit test

A sports director compares form one interests in sporting disciplines with the form two interest distribution, which is Cricket 21.1%, Hockey 27.0%, Rugby 33.9% and Soccer 18.0%. A random sample of 200 form ones gave the frequencies Cricket 42, Hockey 62, Rugby 64 and Soccer 32. State the number of degrees of freedom for the chi-squared goodness of fit test.

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[2 marks]Chi-squared goodness of fit test

A sports director compares form one interests in sporting disciplines with the form two interest distribution, which is Cricket 21.1%, Hockey 27.0%, Rugby 33.9% and Soccer 18.0%. A random sample of 200 form ones gave the frequencies Cricket 42, Hockey 62, Rugby 64 and Soccer 32. The test uses 3 degrees of freedom. State the critical value of the chi-squared statistic at the 5% level of significance.

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[3 marks]Chi-squared goodness of fit test

A sports director compares form one interests in sporting disciplines with the form two interest distribution, which is Cricket 21.1%, Hockey 27.0%, Rugby 33.9% and Soccer 18.0%. A random sample of 200 form ones gave the frequencies Cricket 42, Hockey 62, Rugby 64 and Soccer 32. The chi-squared statistic is 1.844 and the critical value at the 5% level with 3 degrees of freedom is 7.815. What is the conclusion of the test?

A $1.844<7.815$ , so $H_{0}$ is retained: no sufficient evidence of a difference at the 5% level.
B $1.844<7.815$ , so $H_{0}$ is rejected and the form one interests differ from the form two distribution.
C $1.844>7.815$ , so $H_{0}$ is rejected and the form one interests differ from the form two distribution.
DThe test is inconclusive, because one of the expected frequencies falls below 5 and the classes were not pooled.

Section a, Question 11

[3 marks]Regression and correlation

At Kurerana High School the time T, in hours, that each of 8 chemistry students spent studying and the mark M each scored out of 50 were: T = 4, 3, 4, 5, 4, 7, 7, 8 with the corresponding M = 37, 32, 35, 40, 40, 44, 42, 48. For these data

\sum T=42

\sum M=318

\sum T^{2}=244

\sum M^{2}=12\,822

and

\sum TM=1\,730

. Find the gradient

b

of the regression line

M=a+b\text{T}

, giving your answer correct to 3 significant figures.

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[3 marks]Regression and correlation

\sum T=42

\sum M=318

\sum T^{2}=244

\sum M^{2}=12\,822

and

\sum TM=1\,730

. Find the intercept

a

of the regression line

M=a+b\text{T}

, giving your answer correct to 3 significant figures.

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[3 marks]Regression and correlation

\sum T=42

\sum M=318

\sum T^{2}=244

\sum M^{2}=12\,822

and

\sum TM=1\,730

. The regression line is

M=26.234+2.5745\text{T}

. Use it to estimate the study time, in hours and minutes, for a student who scored 41 marks.

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[3 marks]Regression and correlation

\sum T=42

\sum M=318

\sum T^{2}=244

\sum M^{2}=12\,822

and

\sum TM=1\,730

. Find the product moment correlation coefficient, giving your answer correct to 3 significant figures.

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[2 marks]Regression and correlation

For 8 chemistry students the study time T, in hours, and the test mark M, out of 50, gave a product moment correlation coefficient of

r=0.926

. What does this say about the relationship between study time and test mark?

AThere is a strong negative linear relationship: students who studied longer tended to score lower marks.
BThere is almost no linear relationship, since $r$ must exceed $0.99$ before any pattern can be claimed.
CStudying longer causes a higher mark, and $r$ measures how much of the mark the study time caused.
DThere is a strong positive linear relationship between study time and mark.

Statistics Paper 4 November 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

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The answers, and why they are the answers

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