Question

An engineer is going to redesign an ejection seat for an airplane. The seat was designed for pilots weighing between 140 lb and 201 lb. The new population of pilots has normally distributed weights with a mean of 150 lb and a standard deviation of 33.8 lb. a. If a pilot is randomly​ selected, find the probability that his weight is between 140 lb and 201 lb. The probability is approximately nothing. ​(Round to four decimal places as​ needed.) b. If 40 different pilots are randomly​ selected, find the probability that their mean weight is between 140 lb and 201 lb. The probability is approximately nothing. ​(Round to four decimal places as​ needed.) c. When redesigning the ejection​ seat, which probability is more​ relevant? A. Part​ (b) because the seat performance for a sample of pilots is more important. B. Part​ (a) because the seat performance for a single pilot is more important. C. Part​ (b) because the seat performance for a single pilot is more important. D. Part​ (a) because the seat performance for a sample of pilots is more important.

Answer 1

Population mean, µ = 150

Population standard deviation, σ = 33.8

Probability that the weight is between 140 lb and 201 lb, P(140< X <201) =

= P( [(140-150)/33.8 ] < (X-µ)/σ < [(201-150)/34])

= P(-0.30< z <1.51)

= P( z <1.51 ) - P( z <-0.30)

Using excel function:

= NORM.S.DIST(1.51,1) - NORM.S.DIST(-0.30, 1)

= 0.5524

b) Probability that their mean weight is between 140 lb and 201 lb, P(140 < X̅ < 201) =

= P( [(140-150)/(33.8/√40) ] < (X̅ -µ)/(σ/√n) < [(201-150)/(34/√40) ])

= P(-1.87< z <9.54)

= P( z <9.54 ) - P( z <-1.87)

Using excel function:

= NORM.S.DIST(9.54,1) - NORM.S.DIST(-1.87,1)

= 0.9693

c) Answer A. Part​ (b) because the seat performance for a sample of pilots is more important.