Question

High school graduates: Approximately

78%

of freshmen entering public high schools in the United States in

2005

graduated with their class in

2009

. A random sample of

128

freshmen is chosen. Use

(a)Find the mean

μp

.(b)Find the standard deviation

σp

.

(c)Find the probability that less than

91%

of freshmen in the sample graduated.

(d)Find the probability that between

68%

and

83%

of freshmen in the sample graduated.

(e)Find the probability that more than

68%

of freshmen in the sample graduated.

Cumulative Normal Distribution Table as needed. Round your answers to at least four decimal places if necessary.

Answer 1

a)

mean = 0.78

b)

std.deviation = sqrt(0.78 *(1-0.78)/128)
= 0.0366

c)

Here, μ = 0.78, σ = 0.0366 and x = 0.91. We need to compute P(X <= 0.91). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (0.91 - 0.78)/0.0366 = 3.55

Therefore,
P(X <= 0.91) = P(z <= (0.91 - 0.78)/0.0366)
= P(z <= 3.55)
= 0.9998

d)

Here, μ = 0.78, σ = 0.0366, x1 = 0.68 and x2 = 0.83. We need to compute P(0.68<= X <= 0.83). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (0.68 - 0.78)/0.0366 = -2.73
z2 = (0.83 - 0.78)/0.0366 = 1.37

Therefore, we get
P(0.68 <= X <= 0.83) = P((0.83 - 0.78)/0.0366) <= z <= (0.83 - 0.78)/0.0366)
= P(-2.73 <= z <= 1.37) = P(z <= 1.37) - P(z <= -2.73)
= 0.9147 - 0.0032
= 0.9115

e)

Here, μ = 0.78, σ = 0.0366 and x = 0.68. We need to compute P(X >= 0.68). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (0.68 - 0.78)/0.0366 = -2.73

Therefore,
P(X >= 0.68) = P(z <= (0.68 - 0.78)/0.0366)
= P(z >= -2.73)
= 1 - 0.0032 = 0.9968