Answer to Question #261751 in Statistics and Probability for Henloo

Question #261751

Several intelligence tests follow a normal distribution with a mean of 100 and a standard deviation of 15.


a. Determine the percentage of the population that would obtain a score between 95 and 110.

b. For a population of 2,500, how many are expected to have a score above 125?


1
Expert's answer
2021-11-08T15:36:59-0500

a. P(95<X<110)P(95<X<110)

P(95<X<110)=P(9510015<Z<11010015)P(95<X<110)=P(\frac{95-100}{15}<Z<\frac{110-100}{15})

=P(Z<0.6667)P(Z<0.3333)=P(Z<0.6667)-P(Z< -0.3333)

=0.74750.3694=0.3781=0.7475-0.3694=0.3781

Thus, 37.81% of the population would obtain scores between 95 and 110


b. P(X>125)=1P(X<125)P(X>125)=1-P(X<125)

=1P(Z<12510015)=1P(Z<53)=10.9522=0.0478=1-P(Z<\frac{125-100}{15})=1-P(Z<\frac{5}{3})=1-0.9522=0.0478

For a population of 2,500,

2500×0.0478=119.52500\times0.0478=119.5

Thus, 120 people out of 2500 will score above 125.


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