Question

An engineer is going to redesign an ejection seat for an airplane. The seat was designed for pilots weighing between 130 lb and 181 lb. The new population of pilots has normally distributed weights with a mean of 139 lb and a standard deviation of 30.3 lb. a. If a pilot is randomly​ selected, find the probability that his weight is between 130 lb and 181 lb. The probability is approximately . 5339. ​(Round to four decimal places as​ needed.) b. If 35 different pilots are randomly​ selected, find the probability that their mean weight is between 130 lb and 181 lb. The probability is approximately nothing. ​(Round to four decimal places as​ needed.)

Need help with part B!

Answer 1

Solution :

Given that,

mean = = 139

standard deviation = = 30.3

b) n = 35

=   = 139

= / n = 30.3 / 35 = 5.12

P(130 < < 181)  

= P[(130 - 139) /5.12 < ( - ) / < (181 - 139) / 5.12)]

= P(-1.76 < Z < 8.20)

= P(Z < 8.20) - P(Z < -1.76)

Using z table,  

= 1 - 0.0392

= 0.9608