Solution to Exercise 5.54.

  We are given

$$\mu \; = \; E[X_i] \; = \; 2.65 , \quad \sigma \; = \; \sqrt{V[X_i]} \; = \; .85 .$$

Note that we are told that the X_i's are normally distributed, so

$$Z \; = \; \frac{\bar{X}-\mu}{\sigma/\sqrt{n}}$$

is a N(0,1) r.v. for all sample sizes, i.e. no appeal to the CLT is necessary.

(a) Given that n = 25, we have that

$$P[ \bar{X} \le 3.00 ] \; = \; P \left[ \frac{\bar{X}-\mu}{\sigma/\sqrt{n}} \le \frac{3.00-2.65}{.85/\sqrt{25}} \right]$$

$$\; = \; P[ Z \le 2.06 ] \; = \; \Phi(2.06) \; = \; .9803$$

For the second part, we have

$$P[ 2.65 < \bar{X} \le 3.00 ] \; = \; P \left[ \frac{2.65-2... ...u}{\sigma/\sqrt{n}} \le \frac{3.00-2.65}{.85/\sqrt{25}} \right]$$

$$\; = \; P[ 0 < Z \le 2.06 ] \; = \; \Phi(2.06) - \Phi(0) \; = \; .9803 - .5000 \; = \; .4803 .$$

(b) We want to find n so that

$$P[ \bar{X} \le 3.00 ] \; = \; .99 .$$

Now from Table 4.1 on p. 163 we have that

$$P[Z \le 2.33] \; = \; .99 .$$

Thus, we want

$$\frac{3.00-2.65}{.85/\sqrt{n}} \; = \; 2.33 .$$

Algebraic manipulations give

$$\frac{0.35}{2.33} \; = \; .85/\sqrt{n}.$$

$$\frac{2.33}{0.35} \; = \; \sqrt{n}/.85$$

$$n \; = \; \left( 2.33*.85/.35 \right)^2 \; = \; 32.02 .$$

Of course, we can't take n = 32.02 since n must be a whole number. Obviously, n = 32 will get us pretty close, but the problem says ``at least .99'' so we have to take n = 33 since n=32 is slightly too small.