You collect a random sample of 50 children and compute , the average height in inches. The sample standard deviation (s) equals 10. Suppose the true mean height μ equals 45 inches. For which of the following values of will a 95% confidence interval contain the true mean μ? I. = 47.5 inches II. = 50 inches III. = 42 inches

We can use the formula for a confidence interval:

Confidence Interval = \bar{x} \pm z \times \frac{s}{\sqrt{n}},

where:
\bar{x} is the sample mean,
z is the Z-value for the desired confidence level (95% in this case),
s is the sample standard deviation,
n is the sample size.

Given values:
\bar{x} = \mu = 45 inches,
s = 10 inches,
n = 50 children,
For 95% confidence level, the Z-value is 1.96 (you can refer to Z-table for this value).

Plugging in these values we get:

I. CI_{I} = 47.5 \pm 1.96 \times \frac{10}{\sqrt{50}},

II. CI_{II} = 50 \pm 1.96 \times \frac{10}{\sqrt{50}},

III. CI_{III} = 42 \pm 1.96 \times \frac{10}{\sqrt{50}}.

Calculating these confidence intervals gives:

I. CI_{I} = 47.5 \pm 2.78,
II. CI_{II} = 50 \pm 2.78,
III. CI_{III} = 42 \pm 2.78.

Only the 95% confidence interval for I. contains the true mean height 45 inches.

\boxed{\text{Answer: I. } \mu = 47.5 \text{ inches}}