Question

X ~ N(50, 9). Suppose that you form random samples of 25 from this distribution. Let X  be the random variable of averages. Let ΣX be the random variable of sums.

A. Find the 30th percentile. (Round your answer to two decimal places.)

B. find the probability. (Round your answer to four decimal places.)

P(18 < X < 49) =

C. Give the distribution of ΣX.

D. Find the minimum value for the upper quartile for ΣX. (Round your answer to two decimal places.)

E. find the probability. (Round your answer to four decimal places.)

P(1150 < ΣX < 1300)

Answer 1

Answer:

Given,

X ~ (50 , 9)

standard deviation = sqrt(variance) = sqrt(9) = 3

a)

P(X < x) = 0.30

P((x-u)/s < (x - 50)/3) = 0.30

P(z < (x-50)/3) = 0.30

since from z table

P(z < - 0.524) = 0.30

consider,

(x - 50)/3 = - 0.524

x = 48.43

b)

P(18 < X < 49) = P((18 - 50)/3 < (x-u)/s < (49 - 50)/3)

= P(- 10.67 < z < - 0.33)

= P(z < - 0.33) - P(z < - 10.67)

= 0.3707 - 0 [since fro z table]

= 0.3707

c)

ΣX ~ N(50*25 , 9*25) = N(1250 , 225)

standard deviation = sqrt(225) = 15

d)

P(z < a) = 0.75

P(z < (a - 1250)/15) = 0.75

since from standard normal table

P(z < 0.674) = 0.75

so,

(a - 1250)/15 = 0.674

a = 1260.11

e)

P(1150 < Σx < 1300) = P((1150-1250)/15 < z < (1300-1250)/15)

= P(-6.66 < z < 3.33)

= P(z < 3.33) - P(z < - 6.66)

= 0.9995658 - 0.0000033 [since fro z table]

= 0.9996