Question

Assume that human body temperatures are normally distributed with a mean of 98.21°F and a standard deviation of 0.64°F. a. A
healthy people to b. Physicians want to exceed it? (Such a res Click to view page 10 NEGATIVE z Scores (Round to two decimal
k to view page 10 The percentage of sund to two decimal POSITIVE z Scores Standard Normal (2) Distribution: Cumulative Area f
0 0
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Answer #1

SOLUTION:

Solution :

Given that ,

mean = \mu = 98.21

standard deviation = \sigma =0.64

P(x <100.6 ) = P(( x -\mu) / \sigma (100.6-98.21) / 0.64)

= P(z <3.73 )

Using z table

= 0.9999

=99.99%

B.

Using standard normal table,

P(Z > z) = 5%

= 1 - P(Z < z) = 0.05  

= P(Z < z) = 1 - 0.05

= P(Z < z ) = 0.95

= P(Z < 1.65 ) = 0.95  

z =1.65

Using z-score formula,

x = z * \sigma + \mu

x = 1.65 * 0.64+98.21

x = 98.266

=98.27

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