Question

The body temperatures of adults are normally distributed with a mean of 98.60˚F and a standard deviation of 0.73˚F. Step 1 of 4: What temperature would put you in the 77th percentile? Round to 2 decimals. Step 2 of 4: What temperature would put you in the bottom 20% of temperatures? Round to 2 decimals. Step 3 of 4: What is the probability that someone has a body temperature of 100°F or more? Step 4 of 4: What is the probability that someone has a body temperature less than 98°F?

Answer 1

Given,

= 98.6 , = 0.73

a)

77th percentile = + Z * , Where Z is critical value at 77% confidence level.

= 98.6 + 0.7388 * 0.73

= 99.1393

b)

We have to calculate x such that P( X < x) = 0.20

That is we have to calculate 20th percentile.

77th percentile = + Z * , Where Z is critical value at 20% confidence level.

= 98.6 + (-0.8416) * 0.73

= 97.9856

c)

Using standard normal conversion.,

P( X < x) = P (Z < x - / ​​​​​​)

So,

P( X >= 100) = P( Z >= 100 - 98.60 / 0.73)

= P(Z >= 1.9178)

= 0.0276

d)

P( X < 98) = P( Z < 98 - 98.60 / 0.73)

= P( Z < -0.7895)

= 0.2149