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 Σ*X* 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 Σ*X*.

D. Find the minimum value for the upper quartile for
Σ*X*. (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

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