Question

Normal distribution about a mean of 60 seconds and a standard deviation of 8 seconds. If 200 patient were selected at random, how many of them could be taken to the hospital in
(a) Less than 50 seconds
(b) more than 64 seconds

Answer 1

The question seems to be missing the first line.

Let X = the random variable denoting the time taken to reach the hospital

We are given,

X ~ N($\mu$ = 60, $\sigma$² = 8²)

Then we have,

Z = $\frac{X-\mu}{\sigma}$ ~ N(0, 1), Z is the standard normal variate

(a) The probability that a patient could be taken to the hospital in less than 50 seconds

P(X < 50)

= P($\frac{X-60}{8}<\frac{50-60}{8}$)

= P(Z < -1.25)

= $\Phi$(- 1.25) = 0.1056

$\therefore$ If 200 patient were selected at random the number of them could be taken to the hospital in less than 50 seconds = 200 x 0.1056 = 21.12 $\approx$ 21 (rounded to the nearest integer)

Answer: The number of patients could be taken to the hospital in less than 50 seconds is 21.

(b) The probability that a patient could be taken to the hospital in more than 64 seconds

P(X > 64)

= P($\frac{X-60}{8}>\frac{64-60}{8}$)

= P(Z > 0.5)

= 1 - P(Z $\leq$ 0.5)

= 1 - $\Phi$(0.5)

= 1 - 0.6915 = 0.3085

$\therefore$ If 200 patient were selected at random the number of them could be taken to the hospital in more than 64 seconds = 200 x 0.3085 = 61.7 $\approx$ 62 (rounded to the nearest integer)

Answer: The number of patients could be taken to the hospital in more than 64 seconds is 62.