Question

. Find

a. P(Z=-1.01)

b. P(Z>10)

c. P(-1.5<Z<2)

d. P(-2.58<Z)

e. P(Z<3.15)

f. The 30th percentile of Z

g. The z value with 12% area to its right.

2. If X is normally distributed with a mean of 5.6 and a variance of 27, find

a. P(X<2)

b. P(-2.5<X<7.2)

c. The 80th percentile of X

Answer 1

Solution:-

a. P(Z = -1.01) is undefined

b. P(Z > 10) = 0

c. P(-1.5<Z<2) = 0.9104

d. P(-2.58<Z) = 0.9951

e. P(Z < 3.15) = 0.9992

f. The 30 th percentile of Z is -0.5244

g. The z value with 12% area to its right is 1.1750

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2. mean = 5.6 standard deviation = sqrt(variance) = sqrt(27) = 5.1962

a. P(X < 2) = P((x-mu)/sd < (2-5.6)/5.1962)

= P(Z < -0.6928)

= 0.2451

b. P(-2.5 < X < 7.2) = P((-2.5-5.6)/5.1962 < Z < (7.2-5.6)/5.1962)

= P(-1.5588 < Z < 0.3079)

= 0.5623

c. The 80th percentile of X is 9.9731

X = 5.6 + (0.8416*5.1962) = 9.9731

By using Z-score table (left and right)