Question

The mean mass for ostrich eggs is 3.1 pounds, and standard deviation is .5 pounds. The masses

have been found to be normal in distribution. Use this information to answer the following

questions.

(a) Determine the probability that a randomly selected ostrich egg has a mass of more than

3.2 pounds.

(b) Determine probability that 9 randomly selected ostrich eggs have masses with sample

mean more than 3.2 pounds.

(c) For part b, was it necessary to know that the masses were normally distributed, and why

or why not?

(d) Determine probability that a sample of 81 such eggs have mean mass between 3.2 and

3.25 pounds.

(e) For part d, was it necessary to know that the masses were normally distributed, and why

or why not?

(f) Determine egg masses corresponding to top 15%.

Answer 1

Mean, = 3.1 pounds

Standard deviation, = 0.5 pounds

Let X denote the mass of ostrich egg (in pounds)

(a) The required probability = P(X > 3.2)

= P{Z > (3.2 - 3.1)/0.5}

= P(Z > 0.2) = 0.4207

(b) Corresponding to n = 9, standard error = 0.5/√9 = 0.167 pounds

The required probability = P( > 3.2) = P{Z > (3.2 - 3.1)/0.167}

= P(Z > 0.6) = 0.2743

(c) Yes, it was necessary to know that the masses were normally distributed since for n = 9, we cannot assume the sampling distribution of the sample mean to follow Normal distribution if the population is not normaly distributed.

(d) Standard error = 0.5/√81 = 1/18

The required probability = P(3.2 < < 3.25)

= P(1.8 < Z < 2.7) = 0.0324

(e) No, since for a sample size greater than 30, from Central Limit Theorem the sampling distribution of the sample means follows Normal distribution irrespective of the population distribution

(f) Corresponding to top 15% egg masses, the critical z value = 1.0365

Thus, the required egg mass = 3.1 + 1.0365*0.5 = 3.62 pounds