Question

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 138 lb and a standard deviation of 34.9 lb.

a. If a pilot is randomly selected, find the probability that his weight is between 150 lb and 201 lb.
b. If 39 different pilots are randomly selected, find the probability that their mean weight is between 150 lb and 201 lb.
c. When redesigning the ejection seat which probability is more relevant?

Answer 1

Answer:

The answer is below

Step-by-step explanation:

Given that:

mean (μ) = 138 lb, standard deviation (σ) = 34.9 lb

z score is used in statistic to determine by how many standard deviations the raw score is above or below the mean. It is given by:

$z=\frac{x-\mu}{\sigma}$

a) For probability that his weight is between 150 lb and 201 lb, we need to calculate the z score for 150 lb and for 201 lb.

For 150 lb:

$z=\frac{x-\mu}{\sigma}=\frac{150-138}{34.9}=0.34$

For 201 lb:

$z=\frac{x-\mu}{\sigma}=\frac{201-138}{34.9}=1.81$

From normal distribution table, probability that his weight is between 150 lb and 201 lb = P(150 < x < 201) = P(0.34 < z < 1.81) = P(z < 1.81) - P(z < 0.34) = 0.9649 - 0.6331 = 0.3318 = 33.18%

b) If 39 different pilots are randomly selected i.e. n = 39. For probability that his weight is between 150 lb and 201 lb, we need to calculate the z score for 150 lb and for 201 lb.

For 150 lb:

$z=\frac{x-\mu}{\sigma/\sqrt{n} }=\frac{150-138}{34.9/\sqrt{39} }=2.15$

For 201 lb:

$z=\frac{x-\mu}{\sigma/\sqrt{n} }=\frac{201-138}{34.9/\sqrt{39} }=11.3$

From normal distribution table, probability that his weight is between 150 lb and 201 lb = P(150 < x < 201) = P(2.15 < z < 11.3) = P(z < 11.3) - P(z < 2.15) = 1 - 0.9842 = 0.0158 = 1.58%

c) The probability from part C is more important