75% of all Americans live in cities with population greater than 100,000 people. If 40 Americans...

75% of all Americans live in cities with population greater than 100,000 people. If 40 Americans are randomly selected, find the probability that

a. Exactly 29 of them live in cities with population greater than 100,000 people.
b. At most 33 of them live in cities with population greater than 100,000 people.
c. At least 32 of them live in cities with population greater than 100,000 people.
d. Between 26 and 33 (including 26 and 33) of them live in cities with population greater than 100,000 people.

Solutions

Expert Solution

Given:

Probability of success, p = 0.75

Number of sample, n = 40

Let X be the number of Americans live in cities with population greater than 100,000 people.

X follows the Binomial distribution with parameters n = 40 and p = 0.75

X ~ Binomial (n=40, p=0.75)

The probability function of Binomial Distribution is given by

Therefore the probability that

a) Exactly 29 of them live in cities with population greater than 100,000 people is 0.1312

b) At most 33 of them live in cities with population greater than 100,000 people is 0.9038

c) At least 32 of them live in cities with population greater than 100,000 people is 0.2998

d) Between 26 and 33 (including 26 and 33) of them live in cities with population greater than 100,000 people is 0.8493

Answer are rounded to four decimal places

orchestra answered 1 year ago

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