Question

5. Suppose that annual precipitation is uniformly distributed, and ranges between 25 inches and 72 inches.

(a) What is the probability that a year will have between 29 and 44 inches of precipitation? (2)

(b) What is the 30^th percentile of this distribution? (2)

6. Wheat yields per acre are normally distributed, with mean 46 bushels, and standard deviation 10 bushels.

(a) What is the probability that next year’s yield is between 35 and 50 bushels? (2)

(b) What is the 40^th percentile of this distribution? (2)

PLEASE HELP!!!!

Answer 1

Solution :

Given that,

a = 25

b = 72

(A)P(c < x < d) = (d - c) / (b - a)

P(29 < x < 44) = (44 - 29) / (72 - 25)=0.3191

probability=0.3191

(B)

PERCENTILE

= a + (b - a)p

= 25 + (72 - 25)0.30

=39.1

answer=39.1

(6)

Solution :

Given that ,

mean =   = 46

standard deviation = =10

P(35< x <50 ) = P[(35-46) /10 < (x - ) / < (50-46) /10 )]

= P(-1.1 < Z <0.4 )

= P(Z <0.4 ) - P(Z <-1.1 )

Using z table   

= 0.6554-0.1357

probability= 0.5197

B)

Using standard normal table,

P(Z < z) = 40%

= P(Z < z) = 0.40

= P(Z <- 0.25 ) = 0.40

z = - 0.25 Using standard normal z table,

Using z-score formula  

x= z * +

x= -0.25 *10+46

x= 43.5

x=44