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ZIMSEC A Level · 9164/4 · J2016

Statistics Paper 4 June 2016

Questions
39
Total marks
96
Syllabus code
9164/4

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Questions
39
Pass mark
24
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Answer every question in the printed order, get marked at the end, then see the answers.

The questions

Section a

Section a, Question 2

[2 marks]Normal distribution

The diameters of washers produced by a machine follow a Normal distribution with standard deviation 0.1 mm and unknown mean μ\mu. The mean is to be set so that the probability that a diameter exceeds 2.0 mm is 0.03.

Writing the condition as P(Z<2−μ0.1)=0.97P\left(Z < \dfrac{2 - \mu}{0.1}\right) = 0.97, state the value of z for which Φ(z)=0.97\Phi(z) = 0.97.

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[2 marks]Normal distribution

The diameters of washers produced by a machine follow a Normal distribution with standard deviation 0.1 mm. The mean is to be set so that there is a probability of only 3% that a diameter exceeds 2.0 mm.

Given that Φ(1.881)=0.97\Phi(1.881) = 0.97, find the mean diameter in mm.

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[2 marks]Continuous random variables

A continuous random variable X has probability density function f(x) with f(x)=0f(x) = 0 for x<1x < 1 and for x>5x > 5. Between 1 and 5 the density is made of three straight pieces: it is constant at k from x=1x = 1 to x=3x = 3, rises in a straight line from k at x=3x = 3 to 2k at x=4x = 4, then falls in a straight line from 2k at x=4x = 4 to 0 at x=5x = 5.

Which equation in k does the total area under the density give?

  1. Ak+3k2+2k=1k + \dfrac{3k}{2} + 2k = 1
  2. B2k+3k2+k=12k + \dfrac{3k}{2} + k = 1
  3. C2k+3k2+2k=12k + \dfrac{3k}{2} + 2k = 1
  4. D2k+3k+k=12k + 3k + k = 1
[2 marks]Continuous random variables

A continuous random variable X has a probability density function that is constant at 29\dfrac{2}{9} for 1≤x≤31 \le x \le 3, rises in a straight line from 29\dfrac{2}{9} at x=3x = 3 to 49\dfrac{4}{9} at x=4x = 4, then falls in a straight line from 49\dfrac{4}{9} at x=4x = 4 to 0 at x=5x = 5, and is 0 elsewhere.

Which expression gives f(x) on the rising piece 3≤x≤43 \le x \le 4?

  1. Af(x)=29x−49f(x) = \dfrac{2}{9}x - \dfrac{4}{9}
  2. Bf(x)=29x+49f(x) = \dfrac{2}{9}x + \dfrac{4}{9}
  3. Cf(x)=49−29xf(x) = \dfrac{4}{9} - \dfrac{2}{9}x
  4. Df(x)=49x−29f(x) = \dfrac{4}{9}x - \dfrac{2}{9}
[2 marks]Conditional probability

Bag A contains 3 red balls and 2 white balls. Bag B contains 2 red balls and 3 white balls. A bag is selected at random and two balls are drawn from it, one after the other without replacement.

Find the probability that the two balls drawn are red.

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[3 marks]Conditional probability

Bag A contains 3 red balls and 2 white balls. Bag B contains 2 red balls and 3 white balls. A bag is selected at random and two balls are drawn from it, one after the other without replacement.

Given that the two balls drawn are red, find the probability that they came from bag A.

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Section a, Question 3

[2 marks]Discrete random variables

An unbiased tetrahedral die has faces marked 1, 2, 3 and 4. Two such dice are tossed and X is the sum of the two scores.

Find P(X=5)P(X = 5).

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[3 marks]Expectation

Two unbiased tetrahedral dice with faces marked 1, 2, 3 and 4 are tossed and X is the sum of the two scores, so that P(X=x)P(X = x) is 116, 216, 316, 416, 316, 216, 116\dfrac{1}{16},\ \dfrac{2}{16},\ \dfrac{3}{16},\ \dfrac{4}{16},\ \dfrac{3}{16},\ \dfrac{2}{16},\ \dfrac{1}{16} for x=2,3,4,5,6,7,8x = 2, 3, 4, 5, 6, 7, 8 in turn.

A player pays $1 to play. If the sum is 2, 3 or 4 the player wins nothing; if the sum is 5, 6 or 7 the player wins $2; for a sum greater than 7 the player wins $4.

Find the expected gain, in dollars, from each game.

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[2 marks]Median of a data set

A stem and leaf diagram records the number of hectares owned by farmers around a small town, with the key 1/3 meaning 13 hectares:

StemLeaf
01 4 7
11 3 8 9
20 1 2 4 7 8
30 0 2 3 4 5 7 7
42 3 5 7

Find the median number of hectares.

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[2 marks]Quartiles

The number of hectares owned by 25 farmers, arranged in order, is

1, 4, 7, 11, 13, 18, 19, 20, 21, 22, 24, 27, 28, 30, 30, 32, 33, 34, 35, 37, 37, 42, 43, 45, 47.

Using the position 14(n+1)\dfrac{1}{4}(n + 1), find the lower quartile.

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[2 marks]Interquartile range

The number of hectares owned by 25 farmers, arranged in order, is

1, 4, 7, 11, 13, 18, 19, 20, 21, 22, 24, 27, 28, 30, 30, 32, 33, 34, 35, 37, 37, 42, 43, 45, 47.

Using the positions 14(n+1)\dfrac{1}{4}(n + 1) and 34(n+1)\dfrac{3}{4}(n + 1) for the quartiles, find the interquartile range.

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[2 marks]Unbiased estimates

A dozen loaves of bread taken at random from a large batch had masses, in grammes, of

741; 701; 834; 829; 808; 660; 659; 739; 472; 865; 851 and 801.

Assuming the masses come from a Normal distribution, find the unbiased estimate of the population mean, in grammes, correct to 1 decimal place.

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[3 marks]Unbiased estimates

For a random sample of 12 loaves of bread the masses, in grammes, gave ∑x=8960\sum x = 8960 and ∑x2=6 828 956\sum x^{2} = 6\ 828\ 956.

Assuming the masses come from a Normal distribution, find the unbiased estimate of the population variance, correct to 2 decimal places.

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[2 marks]Confidence intervals

A batch of 50 loaves gave a mean mass of 750 grammes, and the population variance of the loaves is known to be 10 816.

Using the table value z=1.64z = 1.64 for 90% confidence, find the upper limit of the 90% confidence interval for the mean mass, in grammes, correct to 2 decimal places.

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Section a, Question 4

[2 marks]Poisson distribution

The number of passengers arriving at a taxi rank per hour has a Poisson distribution with mean 2.

Calculate the probability that in a particular hour no passenger arrives, correct to 3 decimal places.

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[3 marks]Poisson distribution

The number of passengers arriving at a taxi rank per hour has a Poisson distribution with mean 2. At the beginning of an hour there are 4 taxis available for hire, and each taxi takes only one passenger.

Calculate the probability that 4 taxis is an insufficient number for that hour, correct to 4 decimal places.

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[3 marks]Poisson distribution

The number of passengers arriving at a taxi rank per hour has a Poisson distribution with mean 2.

Find the probability that exactly 2 passengers arrive in 2 consecutive hours, correct to 4 decimal places.

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[2 marks]Normal approximation to the binomial
Under what conditions can the Normal distribution be used as an approximation to the Binomial distribution B(n, p)B(n,\ p), where q=1−pq = 1 - p?
  1. AThe number of trials n is large and p equals 0.5, which is what makes the binomial symmetric about np.
  2. BThe number of trials n is small and p is close to 0, so that np stays below 5 while nq remains large.
  3. CThe number of trials n is large and p is very small, so that np is about 5 and nq is close to n.
  4. DThe number of trials n is large and p is such that both np and nq are greater than 5.
[2 marks]Normal approximation to the binomial

A fair coin is tossed 100 times and X is the number of heads obtained, so X∼B(100, 12)X \sim B\left(100,\ \dfrac{1}{2}\right).

The distribution of X is to be approximated by a Normal distribution with the same mean and variance. State the variance of that Normal distribution.

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[3 marks]Normal approximation to the binomial

A fair coin is tossed 100 times and X is the number of heads obtained, so X∼B(100, 12)X \sim B\left(100,\ \dfrac{1}{2}\right), which is approximated by N(50, 25)N(50,\ 25).

Using a continuity correction, calculate the probability that the number of heads obtained is less than 37, correct to 4 decimal places.

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[2 marks]Poisson approximation to the binomial

For a fair coin tossed 100 times, the probability of obtaining fewer than 37 heads is 0.0035. Now 2 000 such coins are each tossed 100 times, and Y is the number of those 2 000 sets that give fewer than 37 heads, so Y∼B(2 000, 0.0035)Y \sim B(2\ 000,\ 0.0035).

Which approximation to the distribution of Y is appropriate, and why?

  1. AA Normal approximation with mean 0.0035 and variance 7, since p is the mean of the approximating distribution.
  2. BA Poisson approximation with mean 7, since n is large, p is small and np=2 000×0.0035=7np = 2\ 000 \times 0.0035 = 7.
  3. CA Normal approximation with mean 7 and variance 7, since 2 000 repetitions is a large number of trials.
  4. DA Poisson approximation with mean 2 000, since the mean of the approximating distribution is the number of trials.
[3 marks]Poisson approximation to the binomial

The number Y of sets, out of 2 000 sets of 100 coin tosses, that give fewer than 37 heads has Y∼B(2 000, 0.0035)Y \sim B(2\ 000,\ 0.0035), which is approximated by a Poisson distribution with mean 7.

Find the probability that fewer than 37 heads are obtained more than 3 times, correct to 3 decimal places.

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[2 marks]Sums of Normal variables

The mass of the contents of a packet of brand M powdered milk is Normally distributed with mean 500 g and standard deviation 8 g. The mass of its packaging is Normally distributed with mean 20 g and standard deviation 2 g, independently of the contents.

State the variance of the total mass of a randomly chosen packet.

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[3 marks]Sums of Normal variables

The mass of the contents of a packet of brand M powdered milk is Normally distributed with mean 500 g and standard deviation 8 g, and the mass of its packaging is Normally distributed with mean 20 g and standard deviation 2 g, independently of the contents.

Find the probability that a randomly chosen packet has a total mass exceeding 525 grammes, correct to 4 decimal places.

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[3 marks]Sums of Normal variables

The mass of the contents of a packet of brand M powdered milk is Normally distributed with mean 500 g and standard deviation 8 g, and packets are independent of one another.

Find the probability that the total mass of the contents of three randomly chosen packets exceeds 1 515 grammes, correct to 4 decimal places.

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[3 marks]Differences of Normal variables

The contents of a packet of brand N powdered milk are Normally distributed with mean 405 g and standard deviation 6 g. The contents of a packet of brand M are Normally distributed with mean 500 g and standard deviation 8 g. All packets are independent.

Let V=(U1+…+U5)−(X1+…+X4)V = (U_{1} + \ldots + U_{5}) - (X_{1} + \ldots + X_{4}), where the U are five brand N packets and the X are four brand M packets. State the variance of V.

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[3 marks]Differences of Normal variables

The contents of a packet of brand N powdered milk are Normally distributed with mean 405 g and standard deviation 6 g, and the contents of a packet of brand M are Normally distributed with mean 500 g and standard deviation 8 g, all packets independent.

Find the probability that the contents of five randomly chosen brand N packets weigh more than the contents of four randomly chosen brand M packets, correct to 4 decimal places.

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Section a, Question 5

[2 marks]Scatter diagrams

For 7 cars of the same make and model, m is the total distance travelled since the car was new, in thousands of km, and d is the distance the car travels on one litre of petrol, in km:

m ( x 1000 km)50100150200300400500
d (km)23222020171612

Which statement best describes the relationship between m and d?

  1. AThere is a weak negative relationship appearing only above 300 thousand km, the first four points being level.
  2. BThere is a strong positive linear relationship: cars that have covered more kilometres go further on a litre.
  3. CThere is a strong negative linear relationship: the further a car has been driven, the less it covers on a litre.
  4. DThere is no relationship worth reporting: the seven points are scattered with no clear direction or trend.
[3 marks]Regression

For 7 cars, m is the total distance travelled since new, in thousands of km, and d is the distance travelled on one litre of petrol, in km:

m ( x 1000 km)50100150200300400500
d (km)23222020171612

The totals are ∑m=1700\sum m = 1700, ∑d=130\sum d = 130, ∑m2=575 000\sum m^{2} = 575\ 000 and ∑md=27 850\sum md = 27\ 850.

Find the gradient of the regression line of d on m, correct to 4 significant figures.

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[3 marks]Regression

For 7 cars, m is the total distance travelled since new, in thousands of km, and d is the distance travelled on one litre of petrol, in km, with ∑m=1700\sum m = 1700 and ∑d=130\sum d = 130.

The regression line of d on m has gradient −0.02295-0.02295. Find its intercept on the d axis, correct to 4 significant figures.

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[3 marks]Regression

For 7 cars of one model, the regression line of d on m is

d=24.15−0.02295m,d = 24.15 - 0.02295m,

where m is the total distance travelled since the car was new, in thousands of km, and d is the distance travelled on one litre of petrol, in km.

Use the line to estimate the distance travelled per litre by a car which has travelled a total distance of 450 000 km. Give your answer in km correct to 3 significant figures.

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[2 marks]Regression

For 7 cars of one model, the regression line of d on m is d=24.15−0.02295md = 24.15 - 0.02295m, where m is the total distance travelled since the car was new, in thousands of km, and d is the distance travelled on one litre of petrol, in km.

What does the gradient −0.02295-0.02295 mean in this context?

  1. AA car that has covered no distance at all since new is expected to travel 0.02295 km on one litre of petrol.
  2. BEvery thousand km added to a car's total distance cuts the distance it covers on one litre by 2.295 km.
  3. CEvery thousand km added to a car's total distance lowers the distance it covers on one litre by 0.02295 km.
  4. DEach extra km a car covers on one litre goes with 0.02295 thousand km less on its total distance travelled.
[2 marks]Grouped data

The maximum heights cleared by 100 male athletes in a high jump competition were grouped as follows, with the class centres used being 163, 168, 173, 178, 183 and 188 cm in turn:

height, h, (cm)frequency
161<h≤165161 < h \le 1654
165<h≤170165 < h \le 17018
170<h≤175170 < h \le 17537
175<h≤180175 < h \le 18026
180<h≤185180 < h \le 18510
185<h≤190185 < h \le 1905

Find the mean height, in cm.

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[3 marks]Grouped data

For the maximum heights cleared by 100 male athletes, grouped with class centres 163, 168, 173, 178, 183 and 188 cm and frequencies 4, 18, 37, 26, 10 and 5, the mean is 174.75 cm and ∑fx2=3 057 075\sum fx^{2} = 3\ 057\ 075.

Find the unbiased estimate of the variance, correct to 2 decimal places.

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[3 marks]Fitting a Normal distribution

The maximum heights cleared by 100 male athletes are to be modelled by a Normal distribution with mean 174.75 cm and standard deviation 5.7899 cm. The expected frequencies for the first three classes are 3.73 for 161<h≤165161 < h \le 165, 16.00 for 165<h≤170165 < h \le 170 and 31.11 for 170<h≤175170 < h \le 175.

Calculate the expected frequency for the class 175<h≤180175 < h \le 180, correct to 2 decimal places.

(You may use Φ(0.043)=0.5172\Phi(0.043) = 0.5172 and Φ(0.907)=0.8178\Phi(0.907) = 0.8178.)

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[2 marks]Fitting a Normal distribution

The maximum heights cleared by 100 male athletes are modelled by a Normal distribution, and the expected frequencies for the six classes are 3.73, 16.00, 31.11, 30.05, 14.39 and z, the last class being 185<h≤190185 < h \le 190 which carries the whole upper tail.

Find z, correct to 2 decimal places.

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[2 marks]Chi-squared goodness of fit

A chi-squared goodness of fit test is to be carried out on 100 athletes' heights in six classes, with observed frequencies 4, 18, 37, 26, 10 and 5 and expected frequencies 3.73, 16.00, 31.11, 30.05, 14.39 and 4.73 respectively.

Which classes must be combined before the test, and why?

  1. AThe first with the second and the last with the fifth, because an expected frequency below 5 makes a class unreliable.
  2. BThe first with the second only, because it is the single class whose observed frequency falls short of the value five.
  3. CNo classes at all, because every one of the six observed frequencies in the table is already at or above five.
  4. DThe three smallest expected frequencies together, because a chi-squared test needs four classes of roughly equal size.
[3 marks]Chi-squared goodness of fit

After pooling, a chi-squared goodness of fit test has four classes with observed frequencies 22, 37, 26 and 15 and expected frequencies 19.73, 31.11, 30.05 and 19.12 respectively.

Calculate the value of χ2=∑(O−E)2E\chi^{2} = \sum \dfrac{(O - E)^{2}}{E}, correct to 2 decimal places.

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[2 marks]Chi-squared goodness of fit

A chi-squared goodness of fit test compares 100 athletes' heights with a Normal model. After pooling there are four classes, and both the mean and the variance of the Normal model were estimated from the data itself. The test statistic is 2.81 and the test is at the 5% level.

What are the degrees of freedom and the conclusion?

  1. A1 degree of freedom and a critical value of 3.841, constraints lost for the total, the mean and the variance; not rejected.
  2. B1 degree of freedom and a critical value of 3.841, with those same three constraints lost; rejected as a poor fit.
  3. C3 degrees of freedom and a critical value of 7.815, one constraint being lost for the total frequency; not rejected.
  4. D2 degrees of freedom and a critical value of 5.991, constraints lost for the total and for the estimated mean; not rejected.

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