According to a recent poll cholesterol levels in healthy adults are right skewed with a mean of 209 mg/dL and a standard deviation of 35 mg/dL. A random sample of 42 healthy adults is selected. Describe the sampling distribution of the sample mean cholesterol level for the sample of 42 adults.
a. |
X̄ ~ N(209, 5.4006) |
|
b. |
X̄ ~ N(209, 35) |
|
c. |
X̄ ~ AN(209, 5.4006) |
|
d. |
X̄ ~ AN(209, 35) |
Solution :
Given that ,
mean = = 209
standard deviation = = 35
n = 42
sample distribution of sample mean is ,
=
= 209
sampling distribution of standard deviation
= / n = 35/ 42
= 5.4006
a. |
X̄ ~ N(209, 5.4006) |
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