The weight of brains from Alzheimer cadavers varies according to a Normal distribution with mean 1077g...
The weight of brains from Alzheimer cadavers varies according to a Normal distribution with mean 1077g and standard deviation 106g. The weight of an Alzheimer-free brain averages 1250 g. What proportion of brains with Alzheimer disease will weigh more than 1250 g?
| a. |
84.94% |
|
| b. |
5.16% |
|
| c. |
94.84% |
|
| d. |
1.63% |
Homework Answers
Solution:
Given in the question
Mean = 1077
Standard deviation = 106
We need to calculate P(X>1250) = 1-P(X<=1250)
Z = (1250-1077)/106 = 1.63
From Z table we found p-value
P(X>1250) = 1- 0.9484 = 0.0516
So it's answer is B. I.e. 5.16%
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