In: Statistics and Probability

An enginner is going to redesign an ejection seat for airplane. the seat was designed for pilots weighing between 150 Ib and 201 Ib . the new population of pilots has normally distributed weights with a mean of 157 Ib and a standard deviation of 29.7 Ib.

b) if 38 different pilots are randomly selected, find the probability that their mean weight is between 150 Ib and 201 Ib.

the probability is approximately ___

Solution :

Given that ,

mean = = 157

standard deviation = = 29.7

P( 150 < x < 201 )

= P[( 150 - 157 ) / 29.7 ) < (x - ) / < ( 201 - 157 ) / 29.7 ) ]

= P( -0.24 < z < 1.48 )

= P(z < 1.48 ) - P(z < -0.24 )

Using z table,

= 0.9306 - 0.4052

= 0.5254

**Probability = 0.5254**

( b )

n = 38

= 157

= / n = 29.7 / 38 = 4.8180

P( 150 < < 201 )

= P[( 150 - 157 ) / 4.8180 < ( - ) / < ( 201 - 157 ) / 4.8180 )]

= P( -1.45 < Z < 9.13 )

= P(Z < 9.13 ) - P(Z < -1.45 )

Using z table,

= 1.0000 - 0.0735

= 0.9265

**Probability = 0.9265**

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