1)
Given data,
μ = 63.6, σ = 2.5, n = 75
x1 = 63
x2 = 65
By applying normal distribution:-
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)
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.
*Please revert if you have any doubts, happy to help you. Thank you.
Assume that the heights of adult Caucasian women have a mean of 63.6 inches and a...
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