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Novosadov [1.4K]

3 years ago

10

An engineer is going to redesign an ejection seat for an airplane. The seat was designed for pilots weighing between 130 lb and

171 lb. the new population of pilots has normally distributed weights with a mean of 137 lb and a standard deviation of 27.7 lb.
A.) if a pilot is randomly selected, find the probability that his weight is between 150 lb and 201 lb.
The probability is approximately__________. (round to four decimal place as needed.)
B.) If 39 different pilots are randomly selected, find the probability that their mean weight is between 150 lb and 201 lb.
The probability is approximately__________. (round to four decimal place as needed.)
C) When redesigning the ejection seat which probability is more relevant
Part A or Part B

2 answers:

____ [38]3 years ago

4 0

Answer:

hi the figures for the weight of the pilots under consideration is wrong the correct weights are 150 lb and 201 lb

0.0665
0.0106
part A

Explanation:

A) probability that a pilot's weight is between 150 lb and 201 lb

i.e P( p 150 ≤ x ≤ p 201 )

p ( 150 ) = $\frac{150 - mean}{standard deviation}$ = $\frac{150 - 137}{27.7}$ = 0.4693

p ( 201 ) = $\frac{201 - mean}{standard deviation}$ = $\frac{201 - 137}{27.7}$ = 2.3104

P ( p(150) ≤ x ≤ p(201) ) = (2.3104 - 04693) / standard deviation

= 1.8411 / 27.7 = 0.0665

B) probability if 39 different pilots are selected randomly and their mean weight is between 150 lb and 201 lb

i.e P( p(150) ≤ x ≤ p(201) )

p ( 150 ) = $\frac{150 - mean }{standard deviation * \sqrt{39} }$ = $\frac{13}{27.7* \sqrt{39} }$ = $\frac{13}{172.9864}$ = 0.0752

p ( 201 ) = $\frac{201-137}{standard deviation * \sqrt{39} }$ = $\frac{64}{172.9864}$ = 0.3700

P ( p(150) ≤ x ≤ p(201) ) = (0.3700 - 0.0752) / 27.7 = 0.0106

part A is considered because an infinite number of pilots are considered

k0ka [10]3 years ago

3 0

Answer:

A.) 0.3088

B.) 0.0017

C.) part A

Explanation:

A.)

$z1=$ $\frac{\left(150-137\right)}{27.7}$ $=0.4693$

$z2=\frac{\left(201-137\right)}{27.7}=2.3105$

$P(0.4693$

B.)

$z1=\frac{150-137}{27.7/ \sqrt{39}} =2.9309\\z2=\frac{201-137}{27.7/ \sqrt{39}}=14.4289$

$\\P(2.9309$

C.) Since the seat performance for an individual pilot is more important than 39 different pilots.

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