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Sampling and the sample mean

Practice section 3 of 5

Questions 13–18 in this collection

Practice questions

Discussion practice · 1 of 6

Separate individual observations from a normal sample mean

The weight of Horlicks in glass jars labelled 500 g is normally distributed with mean 505.1 g and standard deviation 7.5 g. The weight of an empty glass jar is normally distributed with mean 350 g and standard deviation 5.5 g. A glass jar’s weight is independent of the weight of Horlicks it contains.

(i) Find the probability that a randomly selected jar weighs less than 365 g and contains less than 499 g of Horlicks.

(ii) A random sample of 50 jars of Horlicks is taken and the mean Horlicks weight is calculated. (i) Find the probability that this sample mean lies between 505 g and 508 g. (ii) For 100 such samples, each of 50 jars, how many would you expect to have sample mean greater than 504 g?

CJC · Discussion · Question 3

Solution

Discussion practice · 2 of 6

Combine normal observations and choose a minimum sample size

The mass of a Munchi pear is normally distributed with mean 120 g and standard deviation 10 g.

(i) Two pears are chosen at random. Find the probability that one has mass between 100 g and 126 g while the other has mass greater than 130 g.

(ii) Ten pears are chosen at random. Find the probability that exactly three have mass greater than 122 g each.

The mass of a Fuchi apple is normally distributed with mean 115 g and standard deviation 8 g.

(iii) Find the probability that the total mass of three randomly chosen Fuchi apples exceeds three times the mass of one randomly chosen Munchi pear.

(iv) For a random sample of nn Fuchi apples, find the least nn such that the probability that its sample mean differs from 115 g by more than 4 g is not more than 0.3.

[2010/NYJC/II/12]

CJC · Discussion · Question 4

Solution

Discussion practice · 3 of 6

Distinguish CLT estimates from exact probabilities and strict thresholds

(a) A random sample of size 50 is taken from X∼B(10,0.4)X\sim B(10,0.4). Find the probability that the sample mean is less than 4.5.

(b) YY is normally distributed with mean 4 and variance 3, and Yˉ\bar Y is the mean of nn independent observations. (i) Find P(Yˉ<5)P(\bar Y<5) when n=5n=5. (ii) If YY is not normal, state one condition for applying the central limit theorem to calculate this probability. (iii) If instead YY is uniform on (0,12)(0,12), with E(Y)=6E(Y)=6 and Var⁡(Y)=12\operatorname{Var}(Y)=12, find the smallest sample size such that P(Yˉ>5)>0.99P(\bar Y>5)>0.99.

CJC · Discussion · Question 5

Solution

Discussion practice · 4 of 6

Choose a sample size from an exact normal mean distribution

The life, in hours, of a randomly chosen light bulb is normally distributed with mean 1100 and standard deviation 70. How large a sample is required so that the probability that its mean life exceeds 1120 is not more than 5%? Do you need the central limit theorem?

CJC · Discussion · Question 6

Solution

Discussion practice · 5 of 6

Translate between a sample mean and a total

The mean time spent by 12-year-old children watching television in one day is 3.5 hours, with standard deviation 0.7 hours.

(i) For a random sample of 100 children, find the probability that the mean watching time is between 3.4 and 3.7 hours.

(ii) Find the probability that their total Sunday watching time exceeds 365 hours.

CJC · Discussion · Question 7

Solution

Review practice · 6 of 6

Estimate emissions mean and variance from test data

A manufacturer launches the environmentally marketed “Green Leaf” car and claims its carbon emission is at most 80 g/km. The transport authority suspects this figure is understated and requests test data from 20 units:

Carbon emission (g/km)7879808182
Number of units23645

Calculate unbiased estimates of the population mean and variance. [2]

[2016/PJC/II/10(i)]

CJC · Review problems · Question 1

Solution

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