Question

Healty people have body temperatures that are normally distributed with a mean of 98.20 °F and a standard deviation of 0.62 °F .

(a) If a healthy person is randomly selected, what is the probability that he or she has a temperature above 98.8 °F ? answer:

(b) A hospital wants to select a minimum temperature for requiring further medical tests. What should that temperature be, if we want only 2.5 % of healty people to exceed it?

Answer #1

Solution:

Given that,

mean = = 98.20 ^{0}F

standard deviation = = 0.62 ^{0}F

a) p ( x > 98.8 )

= 1 - p (x < 98.8 )

= 1 - p ( x - / ) < ( 98.8 - 98.20 / 0.62)

= 1 - p ( z < 0.6 /0.62 )

= 1 - p ( z < 0.98 )

Using z table

= 1 - 0.8366

= 0.1634

Probability = 0.1634

b )Using standard normal table,

P(Z > z) = 2.5%

1 - P(Z < z) = 0.025

P(Z < z) = 1 - 0.025 = 0.975

P(Z < 1.96) = 0.975

z = 1.96

Using z-score formula,

x = z * +

x = 1.96 * 0.62 + 98.20

= 99.4152

= 99.42^{0}F

Minimum temprature = 99.42^{0}F

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