Solution to Exercise 5.48.

  (a) Given information is:

$$\mu \; = \; E[X_i] \; = \; 50 , \quad \sigma^2 \; = \; V[X_i] \; = \; 1 , \quad \& n \; = \; 100 .$$

Therefore,

$$E[\bar{X}] \; = \; \mu \; = \; 50 ,$$

and

$$V[\bar{X}] \; = \; \sigma^2/n \; = \; 1/100 ,$$

so for the standard deviation (or standard error) of $\bar{X}$ we have

$$\sigma_{\bar{X}} \; = \; \sqrt{1/100} \; = \; .1 .$$

Letting Z denote a N(0,1) r.v. we have by the CLT,

$$P[ 49.75 < \bar{X} < 50.25 ] \; = \; P \left[ \frac{49.75-5... ...ac{\bar{X}-\mu}{\sigma_{\bar{X}}} < \frac{50.25-50}{.1} \right]$$

$$\; \doteq \; P[ -2.5 < Z < 2.5 ] \; = \; \Phi(2.5) - \Phi(-2.5) \; = \; .9938 -.0062 \; = \; .9876 .$$

(b) Changing the above with

$$\mu \; = \; 49.8 ,$$

we have

$$P[ 49.75 < \bar{X} < 50.25 ] \; = \; P \left[ \frac{49.75-4... ...{\bar{X}-\mu}{\sigma_{\bar{X}}} < \frac{50.25-49.8}{.1} \right]$$

$$\; \doteq \; P[ -0.5 < Z < 4.5 ] \; = \; \Phi(4.5) - \Phi(-0.5) \; = \; 1.000-.3085 \; = \; .6115 .$$

Note that Table 3 stops tabulating $\Phi(z)$ at z = 3.49, and $\Phi(3.49) = 0.9998$, so we can safely take $\Phi(4.5) = 1$.