Question

The daily low temperature in Woodland, in December is normally distributed with mean 39.6°F and a standard deviation of 4.1°F. (a) Find the probability that the low temperature on a given day in Woodland, in December, will be less than 32°F. (b) What temperature separates the coldest 4% of all low temperatures in Woodland, in December, from the rest?

Answer 1

Solution :

Given that,

mean = = 39.6

standard deviation = = 4.1

a ) P( x < 32 )

P ( x - / ) < ( 32 - 39.6 / 4.1)

P ( z < -7.1 / 4.1 )

P ( z < -1.73 )

= 0.0418

Probability = 0.0418

b ) P(Z < z) = 4%

= P(Z < z) = 0.04  

= P(Z <-1.751 ) = 0.04

z = -1.75

Using z-score formula,

x = z * +

x = -1.75 * 4.1 +39.6

x = 32.42