Question

VULJUURIU The National Australia Bank undertook a study to monitor teller service times for their surburban Perth branches. They found service times could be adequately approximated by a normal distribution with a mean of 137 seconds and a standard deviation of 29 seconds. Determine the probability that a teller in one of these branches completes their next service in less than two minutes (four decimal places).

Answer 1

Answer:

$\mu$ = 137, $\sigma$ = 29, n=1

P(x< 120) = ?

2 - U 2 = 0/vn

120 - 137 29/1

z = -0.5862

z = -0.59

P(x < 120) = P(z < -0.59)

find P(z < -0.59) using z table we get

P(z < -0.59) = 0.27760

P(x < 120) = P(z < -0.59)

hence ,

P(x < 120) = 0.2776

Probability = 0.2776