Question

According to recent passport data, the percentage of people in a particular country who have a passport has risen dramatically. In 2007, only 25% of the country's citizens had a passport; in 2017 that percentage had risen to 44%. Assume that currently 44% of the citizens have a passport. Suppose 50 citizens of the country are selected at random. Complete parts (a) through (c) below.

a. Find the probability that fewer than 24 have a passport. The probability that fewer than 24 have a passport is nothing. (Type an integer or decimal. Round to three decimal places as needed.)

b. Find the probability that at most 28 have a passport. The probability that at most 28 have a passport is nothing. (Type an integer or decimal. Round to three decimal places as needed.)

c. Find the probability that at least 29 have a passport. The probability that at least 29 have a passport is nothing. (Type an integer or decimal. Round to three decimal places as needed.

Show work. How to get from point a to b

Answer #1

Let X be a Binomial random variable which denotes the number of citizens who have a passport

where n = 50, p = 0.44

np = 22

n(1 - p) = 28

Since both np and n(1 - p) are greater than 10, X can be approximated to Normal distribution with

Mean = np = 22

and standard deviation = = 3.51

i.e. X ~ N(22, 3.51)

(a) The required probability = P(X < 24)

Using correction of continuity, P(X < 24) ≈ P(X < 23.5)

= P{Z < (23.5 - 22)/3.51} = P(Z < 0.427)

= **0.6654**

(b) The required probability = P(X ≤ 28)

Using correction of continuity, P(X ≤ 28) ≈ P(X < 28.5)

= P(Z < 1.852)

= **0.9679**

(c) The required probability = P(X ≥ 29)

= 1 - P(X ≤ 28) = 1 - 0.9679 = **0.0321**

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