Question

Suppose that the time duration of a minor surgery is approximately normally distributed with mean equal to 800 seconds and a standard deviation of 40 seconds. Find the probability that a random sample of 16 surgeries will have average time duration of less than 775 seconds. Use the central limit theorem 9-

Answer 1

$\mu$​​​​​​ = 800 seconds

$\sigma$ = 40 seconds

For sampling distribution of mean, with sample size n

n = 16

$\mu_{\bar{x}}$ = $\mu$​​​​​​ = 800 seconds

$\sigma_{\bar{x}}$ = $\sigma/\sqrt{n}$

= $40/\sqrt{16}$

= 10

P($\bar{x}$ < A) = P(Z < (A - $\mu_{\bar{x}}$)/$\sigma_{\bar{x}}$)

P(average time duration of less than 775 seconds) = P($\bar{x}$ < 775)

= P(Z < (775 - 800)/10)

= P(Z < -2.5)

= 0.0062