Question

The mean yearly snowfall for Boston over the past century is 43.5 inches. The standard deviation for this data is 19.4. Assume that the snowfall amounts are approximately normally distributed. If a random sample of 15 winters are selected at random, what is the probability that they will have a sample mean snowfall amount between 42.5 and 52 inches? Use 4 non-zero decimal places in your calculations. Round z-values to 2 decimal places.

Answer 1

Solution :

Given that ,

mean =   = 43.5

standard deviation = = 19.4

n = 15

= 43.5

=  / n= 19.4/ 15=5.01

P(42.5<     <52 ) = P[(42.5-43.5) / 5.01< ( - ) /   < (52-43.5) /5.01 )]

= P( -0.20< Z <1.70 )

= P(Z <1.70 ) - P(Z <-0.20 )

Using z table

=0.9554 - 0.4207

probability= 0.5347