Question

According to a recent study, more than 900 people move to Florida each day! In fact, data show that only

37% of current Florida residents were born in Florida. Let be the proportion of people who were born in

Florida out of a random sample of 154 Florida residents. Determine the following probabilities. Round

solutions to four decimal places, if necessary

Answer 1

Given:

The proportion of people who were born in Florida out of a random sample of 154 Florida residents is denoted by p.

To find:

Probabilities based on the given data.

Solution:

The number of people who were born in Florida out of a random sample of 154 Florida residents can be modeled by a binomial distribution with n = 154 and probability of success p.

The mean of this distribution is given by μ = np, and the standard deviation is given by σ = sqrt(np(1-p)).

Using the given data, we have p = 0.37, so μ = 1540.37 = 57.18 and σ = sqrt(1540.37*0.63) = 5.17.

(a) P(p > 0.4)

Z-score corresponding to p=0.4 is (0.4-0.37)/0.05 = 0.6

P(p > 0.4) = P(z > 0.6) = 0.2743 (using a standard normal table)

(b) P(50 < X < 65)

X has an approximately normal distribution with mean μ = 57.18 and standard deviation σ = 5.17.

Converting the values to z-scores:

z1 = (50 - 57.18) / 5.17 = -1.39

z2 = (65 - 57.18) / 5.17 = 1.52

P(50 < X < 65) = P(-1.39 < Z < 1.52) = 0.8233 (using a standard normal table)

(c) P(X <= 55 or X >= 60)

P(X <= 55 or X >= 60) = P(X <= 55) + P(X >= 60)

Converting the values to z-scores:

z1 = (55 - 57.18) / 5.17 = -0.43

z2 = (60 - 57.18) / 5.17 = 0.55

P(X <= 55 or X >= 60) = P(Z <= -0.43) + P(Z >= 0.55) = 0.4666 + 0.2902 = 0.7568 (using a standard normal table)