Homework SourseA simple random sample of size n36 is obtained from a popula
A simple random sample of size n=36 is obtained from a population with a mean of 64 and a standard deviation of 18.
a) Describe the sampling distribution of x
b) what is P(x 62.6)?
c) What is P(x 68.7)?
d) What is P (59.8 < x < 65.9)?
Solution
a)
It is bell shaped, with the same mean, u(x) = 64, and standard deviation given by
s(x) = s/sqrt(n) = 18/sqrt(36) = 3.
This is all by central limit theorem.
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b)
We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as
x = critical value = 62.6
u = mean = 64
n = sample size = 36
s = standard deviation = 18
Thus,
z = (x - u) * sqrt(n) / s = -0.466666667
Thus, using a table/technology, the left tailed area of this is
P(z < -0.466666667 ) = 0.320369191 [ANSWER]
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c)
We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as
x = critical value = 68.7
u = mean = 64
n = sample size = 36
s = standard deviation = 18
Thus,
z = (x - u) * sqrt(n) / s = 1.566666667
Thus, using a table/technology, the right tailed area of this is
P(z > 1.566666667 ) = 0.058596313 [ANSWER]
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d)
We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as
x1 = lower bound = 59.8
x2 = upper bound = 65.9
u = mean = 64
n = sample size = 36
s = standard deviation = 18
Thus, the two z scores are
z1 = lower z score = (x1 - u) * sqrt(n) / s = -1.4
z2 = upper z score = (x2 - u) * sqrt(n) / s = 0.633333333
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.080756659
P(z < z2) = 0.736742005
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.655985346 [ANSWER]
