Question

(Sample distributions again) Glucose levels in women without diabetes is normally-distributed with a mean of 91.3 mg/dL and a standard deviation of 12.1 mg/dL. You sample 25 random women without diabetes and find the mean glucose level.

(a) What is the mean of this sampling distribution of the mean?

(b) What is the variance of this sampling distribution of the mean?

(c) What is the standard deviation of this sampling distribution of the mean?

(d) What is the probability that the mean glucose level is less than 90 mg/DL?

(e) What is the probability that the mean glucose level is between 85 mg/Dl and 95 mg/DL?

Answer 1

{ Sampling distribution of sample mean :

If simple random sample of size drawn form normal distribution with mean and variance

then sampling distribution of sample mean is approximately normal with mean and

variance = , Standard deviation    }

We are given,

Glucose levels ( X ) in women without diabetes is normally-distributed with a mean =91.3 mg/dL and a standard deviation = 12.1 mg/dL.

sample size n = 25

a) Mean of this sampling distribution of the mean is,

b) Variance of this sampling distribution of the mean is

c) Standard deviation of this sampling distribution of the mean is

d) We have to find P( < 90 )

Using Excel function , =NORMDIST( x , mean , SD , 1 )

P( < 90 ) = NORMDIST( 90 , 91.3 , 2.42 , 1 ) = 0.2956

Probability that the mean glucose level is less than 90 mg/DL is 0.2956.

e ) We have to find P( 85 < < 95 )

P( 85 < < 95 ) = P( < 95 ) - P( < 85 )

Using Excel function , =NORMDIST( x , mean , SD , 1 )

P( < 95 ) = NORMDIST( 95 , 91.3 , 2.42 , 1 ) = 0.936859

P( < 85 ) = NORMDIST( 85 , 91.3 , 2.42 , 1 ) = 0.004616

So, P( 85 < < 95 ) = 0.936859 - 0.004616 =0.9322

Probability that the mean glucose level is between 85 mg/Dl and 95 mg/DL is 0.9322