SOLUTION: <<>>A set of wrist watch prices are normally distributed with a mean of 76 dollars and a standard deviation of 10 dollars. What proportion of wrist watch prices are between 63 doll

Question 1038752: <<>>A set of wrist watch prices are normally distributed with a mean of 76 dollars and a standard deviation of 10 dollars. What proportion of wrist watch prices are between 63 dollars and 90 dollars?<<>>
I believe this is a z-score problem, so I'd be using the equation (X-mean)/standard deviation
... I was thinking since it asks for the por potion between 63 and 90 to take the sum & divide by 2.....
[76-((63+90)/2)]/10
Thank you for your help!!

Found 2 solutions by solver91311, jim_thompson5910:
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!

No.

Find the Z-score for the upper bound and look up the probability of X being less than or equal to that Z-score.

Find the Z-score for the lower bound and look up that probability.

Subtract the lower bound probability from the upper bound probability.

John

My calculator said it, I believe it, that settles it

Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!

You'll need to calculate two z scores.

For the raw score of x = 63, the z score is...

z = (x - mu)/sigma
z = (x - 76)/10
z = (63 - 76)/10
z = -13/10
z = -1.30

For the raw score of x = 90, the z score is...

z = (x - mu)/sigma
z = (x - 76)/10
z = (90 - 76)/10
z = 14/10
z = 1.4

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So we need to find the area under the standard normal bell curve (Z distribution curve) between z = -1.3 and z = 1.4

First find the area to the left of z = -1.3

Use this table locate the row that starts with "-1.3" (page 1) and the column that has 0.00 at the top. The row and column intersect at the value 0.0968

So this means P(Z < -1.3) = 0.0968
Basically, the area to the left of z = -1.3 is 0.0968. Let's make M = 0.0968. We'll use this value of M later.

Now use the table again to find that
P(Z < 1.4) = 0.9192
this value of 0.9192 is found by looking at the row that starts with 1.4 (page 2) and has 0.00 at the top of the column.
Let's make N = 0.9192

Now subtract the values of N and M (big minus small)
N - M = 0.9192 - 0.0968 = 0.8224

So the area under the curve between z = -1.3 and z = 1.4 is 0.8224

This means the proportion of those between 63 dollars and 90 dollars is 0.8224 (which is equivalent to roughly 82.24%). This proportion is approximate because the table values are approximate.

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