Question

Given a standardized normal distribution​ (with a mean of 0 and a standard deviation of​ 1)

(round to 4 decimal places)

a) What is the probability that Z is between −1.59 and 1.88?

b) What is the probability that Z is less than −1.59 or greater than 1.88​?

c) What is the value of Z if only 1​% of all possible Z values are​ larger?

d) Between what two values of Z​ (symmetrically distributed around the​ mean) will 98.36​% of all possible Z values be​contained?

Answer 1

a)

P(-1.59 < Z < 1.88) = P(Z < 1.88) - P(Z < -1.59)

= 0.9699 - 0.0559

= 0.9140

b)

P(z < -1.59 O z > 1.88) = 1 - P(-1.59 < Z < 1.88)

= 1 - 0.9140

= 0.0860

c)

We have to calculate z such that

P(Z > z) = 0.01

P(Z < z) = 1 - 0.01

P(Z < z) = 0.99

From Z table, z-score for the probability of 0.99 is 2.33

z = 2.33

d)

P(-z < Z < z) = 0.9836

P(Z < z) - P(Z < -z) = 0.9836

P(Z < z) - ( 1 - P(Z < z) ) = 0.9836

P(Z < z) - 1 + P(Z < z) = 0.9836

2 P(Z < z) = 1.9836

P(Z < z) = 0.9918

From Z table, z-score for the probability of 0.9918 is 2.40

Two z values are -2.40 and 2.40