91Ó°ÊÓ

We are given 1000 confidence intervals for estimating a population mean \( \mu \). Each interval is a 95% confidence interval. This means that each interval has a probability of 0.95 of capturing the true parameter \( \mu \). We are tasked to find how many intervals we expect to capture \( \mu \) and the probability that between 940 and 960 of these intervals do contain \( \mu \).

Let \( Y \) be the count of intervals that actually contain the true value of \( \mu \). \( Y \) can be modeled as a binomial random variable because each interval either contains \( \mu \) or it doesn't, independently of the other intervals. Specifically, \( Y \sim \text{Binomial}(1000, 0.95) \), where 1000 is the total number of intervals and 0.95 is the probability that a single interval contains \( \mu \).

The expected value \( E(Y) \) for a binomial distribution \( \text{Binomial}(n, p) \) is given by \( np \). Here, \( n = 1000 \) and \( p = 0.95 \), thus:\[ E(Y) = 1000 \times 0.95 = 950 \] Therefore, we expect 950 intervals to capture \( \mu \).

We need to find the probability that \( Y \) lies between 940 and 960. Since \( n \) is large, we can use a normal approximation for the binomial distribution. The mean is \( \mu = np = 950 \) and the standard deviation is \( \sigma = \sqrt{np(1-p)} = \sqrt{1000 \times 0.95 \times 0.05} = \sqrt{47.5} \approx 6.89 \).Using the normal approximation:\[ P(940 \leq Y \leq 960) \approx P\left( \frac{939.5 - 950}{6.89} \leq Z \leq \frac{960.5 - 950}{6.89} \right) \]Calculate the Z-scores:\[ Z_{lower} = \frac{939.5 - 950}{6.89} \approx -1.52 \] \[ Z_{upper} = \frac{960.5 - 950}{6.89} \approx 1.52 \] Look these Z-scores up in the standard normal distribution table or use a calculator to find:\[ P(-1.52 \leq Z \leq 1.52) \approx 0.868 \]

Our calculations estimate that, on average, 950 of the 1000 intervals will contain \( \mu \). The probability that between 940 and 960 of these intervals contain \( \mu \) is approximately 0.868.