Question

Assume that the heights of adult Caucasian women have a mean of 63.6 inches and a standard deviation of 2.5 inches. If 75 women are randomly selected, find the probability that they have a mean height between 63 and 65 inches. O A. 0.3071 B. 0.2119 OC. 0.0188 OD. 0.9811 O E. There is not enough information to determine the probability

Answer 1

1)

Given data,

μ = 63.6, σ = 2.5, n = 75

x₁ = 63

x₂ = 65

By applying normal distribution:-

X-

z(x₁ = 63) = - 2.078 = -2.08

z(x₂ = 65) = 4.85

P( - 2.08 < z < 4.85) = P(z <4.85) - P(z< -2.08) = 0.9999 - 0.0188 = 0.9811

Option D.

2)

Given data,

Sample mean = 5.74 , Standard error of the estimate = 0.20

Confidence Interval = Mean +/- Standard error of the estimate = 5.74+/- 0.20 = (5.54, 5.94)

Option A.

3)

Given data,

Sample size n =3317

Mean = 0.508

Standard deviation = 0.012

A sample proportion from the exit poll of p=0.468 be a plausible value expected in the exit poll. For this we find a confidence interval

= 0.508 +/- 3*0.012 = 0.508 +/- 0.036 = (0.472 , 0.544)

u +3*0

Up to three standard deviation, the proportion is not lies in this interval. So, No. It does not lie within three standard deviations of the mean sample proportion.

No, a sample proportion would not from the exit poll of 0.468 be a plausible value expected in the exit poll.

Option B.

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