Question

- Assume that the variable Z is distributed as a standardized normal with a mean of 0 and a standard deviation of 1. Calculate the following:

a. Pr (Z > 1.06)

b. Pr (Z < - 0.29)

c. Pr (-1.96 < Z < -0.29)

d. The value of Z for which only 8.08% of all possible values of Z are larger than.

- Assume that the variable X is distributed as normal with μ= 51 and σ = 4. Calculate the following:

a. Pr (X = 51)

b. Pr (X > 46)

c. Pr (X < 43)

d. Pr ( 43 < X < 46)

e. Two values of X, symmetric around the mean, between which 70% of all the possible values of X lie.

Answer 1

Q.1)

a) Pr(Z > 1.06) = 1 - Pr(Z < 1.06) = 1 - 0.8554 = 0.1446

Pr(Z > 1.06) = 0.1446

b) Pr(Z < -0.29) = 0.3859

c) Pr(-1.96 < Z < -0.29)

= Pr(Z < -0.29) - Pr(Z < -1.96)

= 0.3859 - 0.0250

= 0.3609

=> Pr(-1.96 < Z < -0.29) = 0.3609

d) we want to find, the z-score such that,

P(Z > z) = 0.0808

=> 1 - P(Z < z) = 0.0808

=> P(Z < z) = 0.9192

Using the standard normal z-table we get, z-score correponds to probability 0.9192 is, z = 1.40

z-score = 1.40

Q.2) Assume that the variable X is distributed as normal with μ= 51 and σ = 4.

a) Pr(X = 50) = 0.0000

b)

Pr(X > 46) = 0.8944

c)

Pr(X < 43) = 0.0228

d)

Pr(43 < X < 46) = 0.082

e) We want to find, the values of x1 and x2 such that,

Pr(x1 < X < x2) = 0.70

First we find, the z-scores such that,

Pr(-z < Z < z) = 0.70

=> 2 * Pr(Z < z) - 1 = 0.70

=> Pr(Z < z) = 0.15

Using standard normal z-table we get, z-score corresponds to probability 0.15 is, z = -1.04

=> Pr(-1.04 < Z < 1.04) = 0.70

Therefore,

x1 = ( Z * σ) + μ = ( - 1.04 * 4 ) + 51 = -4.16 + 51 = 46.84

x2 = ( 1.04 * 4) + 51 = 4.16 + 51 = 55.16

=> (46.84 < X < 55.16) = 0.70

Hence, values of X are, 46.84 and 55.16