Question

-. For what follows, Z denotes the standard normal random variable.

(1) P(Z<1.26) is about

(a) 0.7924

(b) 0.3962

(c) 0.1038

(d) 0.6038

(e) 0.8962

(2) P(-0.54 < Z < 0.78) is about

(a) 0.2054

(b) 0.0769

(c) 0.4877

(d) 0.2823

(e) 0.6877

-Assume the random variable X is normally distributed with mean 172 and standard deviation 10. Find the following probabilities.

(3) P(X<160)

(a) 0.8849

(b) 0.1151

(c) 0.9999

(d) 0.8078

(e) 0.8438

(4) P(X>150)

(a) 0.0139

(b) 0.9861

(c) 0.7910

(d) 0.8599

(e) 0.209

(5) P(X>170 or X<140)

(a) 0.4214

(b) 0.9993

(c) 0.58

(d) none of the above

Answer 1

1)

From Z score table

P(Z<1.26) = 0.8962

Option e is correct

2)

P(-0.54 < Z < 0.78) = P ( z < 0.78) - P( -0.54 < z)

= 0.7823 - 0.2946

= 0.4877

Option C is correct

3)

Mean = 172

S.D. = 10

P(X<160)

From Z score table

Z = (X - ?) / ?
Z = (160 - 172) / 10

Z = -1.2

P(X<160) = P( Z < -1.2) = 0.1151

option B

4)

Z score at 150

Z = (X - ?) / ?
Z = (150 - 172) / 10

Z = -2.2

P(X>150) = P ( Z > -2.2) = 0.9861?

option B

5)

Z score at x=170

Z = (X - ?) / ?
Z = (170 - 172) / 10

Z = -0.2

Z score at x=140

Z = (X - ?) / ?
Z = (140 - 172) / 10

Z = -3.2

P(X>170 or X<140) = P(X > -0.2 or X < -3.2)

= P( X > -0.2) + P( X < -3.2 )

= 0.5793? + 0.0007?

= 0.58