# Statistics - normal distribution question?

The following is the solution to a homework problem - I am having difficulty understanding exactly how the problem was solved because the solution does not go into enough detail. If you could provide a clearer explanation, that would be much appreciated. In addition, one specific question I have is I don't...
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The following is the solution to a homework problem - I am having difficulty understanding exactly how the problem was solved because the solution does not go into enough detail. If you could provide a clearer explanation, that would be much appreciated. In addition, one specific question I have is I don't understand how the sample mean of 16 was calculated for question a) - it does not look like there is enough information to work this out. How was this calculated?

PS I had to replace some mathematical symbols with words as I could not reproduce them here.

The Hamilton (HAM-D) Scale is used to assess patients for clinical depression. In clinically depressed patients, the scale is normally distributed with a mean of 19 points, and a standard deviation of 4 points.

a. Borderline patients have HAM-D scores between 7 and 10. What proportion of depressed patients fall in this range?

Solution:

X=HAM-D score and we know X~N(19,16)

We want P(7<X<10).

Transforming we get:

P(7-19/ √16 < x- mu/standard deviation < 10-19/√16) = P(-3<Z<-2.25)

b. The cutoff for severe depression is the 90th percentile; what raw score does this correspond to on the HAM-D scale?

Solution:

90th percentile in the Z distribution corresponds to 1.282.

z=x-mu/standard deviation=1.282. Solving for x we get:

x=(1.282*4)+19=24.128

c. Consider a random sample of 16 patients. What is the distribution of the sample mean for this sample?

Solution:

x̄ ~ N(19, 16/16)

x̄~ N(19,1)

e. What is the probability that the sample mean of the 16 patients is between 17 and 20?

Solution:

P(17-19/1 < x- mu/variance /n < 20-19/1) = P(-2<Z< 1)

=.841-.023=.8186

PS I had to replace some mathematical symbols with words as I could not reproduce them here.

The Hamilton (HAM-D) Scale is used to assess patients for clinical depression. In clinically depressed patients, the scale is normally distributed with a mean of 19 points, and a standard deviation of 4 points.

a. Borderline patients have HAM-D scores between 7 and 10. What proportion of depressed patients fall in this range?

Solution:

X=HAM-D score and we know X~N(19,16)

We want P(7<X<10).

Transforming we get:

P(7-19/ √16 < x- mu/standard deviation < 10-19/√16) = P(-3<Z<-2.25)

b. The cutoff for severe depression is the 90th percentile; what raw score does this correspond to on the HAM-D scale?

Solution:

90th percentile in the Z distribution corresponds to 1.282.

z=x-mu/standard deviation=1.282. Solving for x we get:

x=(1.282*4)+19=24.128

c. Consider a random sample of 16 patients. What is the distribution of the sample mean for this sample?

Solution:

x̄ ~ N(19, 16/16)

x̄~ N(19,1)

e. What is the probability that the sample mean of the 16 patients is between 17 and 20?

Solution:

P(17-19/1 < x- mu/variance /n < 20-19/1) = P(-2<Z< 1)

=.841-.023=.8186

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