Question

Kelowna is a city on Okanagan Lake in the Okanagan Valley in the southern interior of British Columbia, Canada. The average snowfall in Kelowna is 35 cm per year with a standard deviation of 10 cm. If the given data follows a normal model, what values should border the middle 95% of the model?

Select one:

a. 16.0 and 54.0

b. 15.5 and 45.0

c. 15.3 and 54.6

d. 15.4 and 54.6

e. 27.2 and 36.0

Answer 1

Solution:-

Given that,

mean = = 35

standard deviation = = 10

Using standard normal table,

P( -z < Z < z) = 95%

= P(Z < z) - P(Z <-z ) = 0.95

= 2P(Z < z) - 1 = 0.95

= 2P(Z < z) = 1 + 0.95

= P(Z < z) = 1.95 / 2

= P(Z < z) = 0.975

= P(Z < 1.96 ) = 0.975

= z  ± 1.96

Using z-score formula,

x = z * +

x = -1.96 * 10 + 35

x = 15.4

Using z-score formula,

x = z * +

x = 1.96 * 10 + 35

x = 54.6

The middle 95% are from 15.4 and 54.6.

correct option is = d