Question

A) The weights of 1000 students are normally distributed with a mean of 50 kilograms and a standard deviation of 5 kilograms, If 100 random sample each of size 30 are drawn from this population, determine:
1) The sampling distribution of the average weight of the student
2) The number of sample that fall betweeen 48 and 53 kilograms

Answer 1

1) From the central limit theorem we see that the sampling distribution of the mean approaches a normal distribution with a mean of

$$E(\bar{X})=\mu_{\bar{X}}=\mu=50 \ kg$$

and a variance of

$${\sigma^2 \over n}={5^2 \over 30}={5 \over 6}$$

The standard deviation of the sampling distribution

$$\sigma_{\bar{X}}={5 \over \sqrt{30}}={\sqrt{30}\over 6}$$

2)

$$P(48<\bar{X}<53)=P\bigg({48-50 \over 6/\sqrt{30}}<Z<{53-50 \over 6/\sqrt{30}}\bigg)=$$ $$=P\bigg(Z<{\sqrt{30} \over 2}\bigg)-\bigg(1-P\bigg(Z<{\sqrt{30} \over 3}\bigg)\bigg)\approx$$ $$\approx0.9969-0.0339=0.9630$$

The number of sample that fall betweeen 48 and 53 kilograms

$$100\cdot0.9630=96.3\approx96$$