Answer:
The probability is approximately 0.3193
Step-by-step explanation:
The question parameters;
The mass between of the pilot for which the seat was designed, [tex]\overline x[/tex] = 140 lb and 191 lb
The mean weight of the new pilots, μ = 148 lb
The standard deviation of the weights new pilots, s = 30.3 lb
a. The z-score is given as follows;
[tex]z=\dfrac{\bar{x}-\mu }{{\sigma }}[/tex]
Therefore, we have;
[tex]z_{140}=\dfrac{140-148 }{{30.3 }} \approx 0.264\overline {0264}[/tex]
The p-value = 0.60257
[tex]z_{191}=\dfrac{191-148 }{{30.3 }} \approx 1.419\overline {1419}[/tex]
The p-value = 0.9222
The probability that the mean lie between the two values = 0.9222 - 0.60257 = 0.31963
Therefore, the probability that the weight is between the 140 lb and 191 lb ≈ 0.3196
