7.28.) A sample of 150 is drawn from a population with a proportion equal to 0.40 and a population size of 400. Determine the probability of observing. a.) between 50 and 54 successes. b.) between 55 and 62 successes. c.) between 53 and 70 successes.

a) P(50 < x < 54) = 0.112

b) P(55 < x < 62) = 0.428

c) P(53 < x < 70) = 0.644

Step-by-step explanation:

Sample mean = nP = (sample size) × (population proportion) = 150 × 0.4 = 60

Sample standard deviation = √[nP(1-P)] = √(150×0.4×0.6) = 6

To check if this distribution approximates a normal distribution

np = 60 > 10

np(1-p) = 36 > 10

So, this is a normal distribution problem

a) P(50 < x < 54)

We standardize 50 and 54.

The standardized score for any value is the value minus the mean then divided by the standard deviation.

For 50

z = (x - μ)/σ = (50 - 60)/6 = - 1.67

For 54

z = (x - μ)/σ = (54 - 60)/6 = - 1.00

P(50 < x < 54) = P(-1.67 < z < -1.00) = P(z < -1.00) - P(z < -1.67) = 0.159 - 0.047 = 0.11

b) P(55 < x < 62)

We standardize 55 and 62.

The standardized score for any value is the value minus the mean then divided by the standard deviation.

For 55

z = (x - μ)/σ = (55 - 60)/6 = - 0.83

For 62

z = (x - μ)/σ = (54 - 60)/6 = 0.33

P(55 < x < 62) = P(-0.83 < z < 0.33) = P(z < 0.33) - P(z < -0.83) = 0.630 - 0.202 = 0.428

c) P(53 < x < 70)

P(53 < x < 70)

We standardize 53 and 70.

The standardized score for any value is the value minus the mean then divided by the standard deviation.

For 53

z = (x - μ)/σ = (53 - 60)/6 = - 0.50

For 70

z = (x - μ)/σ = (70 - 60)/6 = 1.67

P(53 < x < 70) = P(-0.50 < z < 1.67) = P(z < 1.67) - P(z < -0.50) = 0.953 - 0.309 = 0.644

Hope this Helps!!!