Question

The body temperatures of adults are normally distributed with a mean of 98.60˚F and a standard deviation of 0.73˚F.

A)What temperature would put you in the 76th percentile? Round to 2 decimals

B)What temperature would put you in the bottom 20% of temperatures? Round to 2 decimals.

C)What is the probability that someone has a body temperature of 100°F or more? [Remember to round probabilities to 4 decimal places]

D)What is the probability that someone has a body temperature less than 96°F?

Please Show step by step.

Answer 1

Mean = 98.60 ^oF

Standard deviation = 0.73 ^oF

P(X < A) = P(Z < (A - mean)/standard deviation)

A) Let P be the 76th percentile

P(X < P) = 0.76

P(Z < (P - 98.6)/0.73) = 0.76

Take Z value corresponding to 0.7600 from standard nrmal distribution table

(P - 98.6)/0.73 = 0.71

P = 99.12 ^oF

So, the 76th percentile is 99.12 ^oF

B) Let T be the 20th percentile

P(X < T) = 0.20

P(Z < (T - 98.6)/0.73) = 0.20

Take Z value corresponding to 0.2000 from standard nrmal distribution table

(T - 98.6)/0.73 = -0.84

T = 97.99 ^oF

So, a temperature below 97.99 ^oF will put you in bottom 20%

C) P(someone has a body temperature of 100°F or more) = P(X > 100)

= 1 - P(X < 100)

= 1 - P(Z < (100 - 98.6)/0.73)

= 1 - P(Z < 1.92)

= 1 - 0.9726

= 0.0274

D) P(someone has a body temperature less than 96°F) = P(X < 96)

= P(Z < (96 - 98.6)/0.73)

= P(Z < -3.56)

= 0.0002