Worked solution
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
Answer
(i) ; (ii)(i) ; (ii)(ii) expected 85 samples.
(i) Independent component weights
Let be the Horlicks weight and the empty-glass weight, in grams. Interpret “jar weighs” as the empty-glass component, as in the source working. Independence gives , to 3 s.f.
(ii)(i) Distribution of the sample mean
For 50 independent jars, exactly because the population is normal. Thus , to 3 s.f. The second parameter is variance; use as the standard deviation in a normal-CDF calculation.
(ii)(ii) Expected number of successful samples
. The expected number among 100 samples is , giving 85 to the nearest integer, as printed.
Common mistakes
Reduce variance by 50 for a sample mean, not standard deviation by 50. The first event uses the empty jar, not the filled jar’s total weight.
Solution notes
No marks are printed for this question; none are invented.
The prompt repeats roman labels: first(i), then(ii) with nested(i),(ii); the source working uses(a),(b)(i),(b)(ii). All three asks are retained. “Jar weighs less than 365 g” is interpreted as empty-glass mass consistently with the source’s explicit J definition; the original wording is preserved in the native record.