Question

The annual rainfall in a certain region is approximately normally distributed with mean 41.2 inches and standard deviation 5.5 inches. Round answers to at least 4 decimal places.

a) What is the probability that an annual rainfall of less than 44 inches?

b) What is the probability that an annual rainfall of more than 39 inches?

c) What is the probability that an annual rainfall of between 38 inches and 42 inches?

d) What is the annual rainfall amount for the 75th percentile?

Answer 1

Solution :

Given that ,

mean = = 41.2

standard deviation = = 5.5

(a)

P(x < 44) = P[(x - ) / < (44 - 41.2) / 5.5]

= P(z < 0.51)

= 0.6950

(b)

P(x > 39) = 1 - P(x < 39)

= 1 - P[(x - ) / < (39 - 41.2) / 5.5]

= 1 - P(z < -0.4)

= 0.6554

(c)

P(38 < x < 42) = P[(38 - 41.2)/ 5.5) < (x - ) /  < (42 - 41.2) / 5.5) ]

= P(-0.58 < z < 0.15)

= P(z < 0.15) - P(z < -0.58)

= 0.5396 - 0.2810

= 0.2586

(d)

P(Z < 0.67) = 0.75

z = 0.67

Using z-score formula,

x = z * +

x = 0.67 * 5.5 + 41.2 = 44.885

75th percentile = 44.885