Question

Given a standardized normal distribution (with a mean of 0 and a standard deviation of 1), what is the probability that:

a. Z is between -1.57 and 1.84

b. Z is less than -1.57 or greater than 1.84

c. What is the value of Z if only 25 percent of all possible Z values are larger

d. Between what two values of Z (symmetrically distributed around the mean) will 68.26 percent of all possible Z values be contained?

Answer 1

Solution:

Z ~ N(0, 1)

a) We have to obtain P(-1.57 < Z < 1.84)

P(-1.57 < Z < 1.84) = P(Z < 1.84) - P(Z < -1.57)

Using "NORM.S.DIST" function of excel we get,

P(Z < 1.84) = 0.9671 and P(Z < -1.57) = 0.0582

Hence, P(-1.57 < Z < 1.84) = (0.9671 - 0.0582) = 0.9089

Probability that Z is between -1.57 and 1.84 is 0.9089.

b) We have to obtain P(Z < -1.57 or Z > 1.84)

P(Z < -1.57 or Z > 1.84) = P(Z < -1.57) + P(Z > 1.84)

Using "NORM.S.DIST" function of excel we get,

P(Z > 1.84) = 0.0329 and P(Z < -1.57) = 0.0582

Hence, P(Z < -1.57 or Z > 1.84) = (0.0582 + 0.0329) = 0.0911

Probability that Z is less than -1.57 or greater than 1.84 is 0.0911.

c) We have to obtain value of Z such that only 25 percent of all possible Z values are larger than this value.

Let the value of Z is b such that only 25 percent of all possible Z values are larger than this value.

Hence, P(Z > b) = 0.25 .........................(1)

Using "qnorm" function of R software we get,

P(Z > 0.67449) = 0.25 ........................(2)

Comparing (1) and (2) we get,

b = 0.67449

The value of Z is 0.67449.

d) According to empirical rule of normal distribution 68.26% of all possible observations fall between (μ- σ) and (μ- σ).

Where, μ is mean and σ is standard deviation.

Z is a standardized normal distribution. Mean of a standard normal distribution is equal to 0 and standard deviation is equal to 1.

i.e.  μ = 0 and σ = 1

Hence,  (μ - σ) = (0 - 1) = 1

and  (μ + σ) = (0 + 1) = 1

Hence, between -1 and 1 (symmetrically distributed around the mean), 68.26% of all possible Z values contained.

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