Question

Assume Z is a random variable with a standard normal distribution and d is a positive number. If P(Z>d)=0.15, then P(−d<Z<d)=0.7

.

    __True
    __False

Please explain, step by step on how to figure this out.

Answer 1

Answer: True

Explanation:

Method - 1

given that,

P(Z>d)=0.15

P(Z < d)= 1 - 0.15

P(Z < d)= 0.85

we want to prove that P(−d<Z<d)=0.7

We know that ,

P(−d<Z<d)= P(z < d ) - P( z < -d )

Where,

P( z < d ) = 0.85

P(z < -d ) =  P(Z>d)= 0.15

P(−d<Z<d)= 0.85 - 0.15

P(−d<Z<d)= 0.70

hence,

given statement is true.

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Method -2

here,  P( Z>d )=0.15

P(Z<d)= 1 -  P(Z>d)

P(Z<d)= 1 - 0.15

P(Z<d)= 0.85

now using normal z table find z-score for 0.85

we get,

z-score = 1.04

thus we get,

d = 1.04

we have also given,

P(−d<Z<d)=0.7

now put value of d into above as follows,

P( -1.04 < Z < 1.04 ) = 0.7

where,

P( -1.04 < Z < 1.04 ) = P(Z < 1.04 ) - P(Z < -1.04 )

using normal z table find the following probabilities.

P(Z < 1.04 ) = 0.8508

P(Z < -1.04 ) = 0.1492

P( -1.04 < Z < 1.04 ) = 0.8508 - 0.1492

P( -1.04 < Z < 1.04 ) = 0.7016

thus we find that ,

P( -1.04 < Z < 1.04 ) = 0.7

hence,

given statement is true.