Question

Let Z be a standard normal random variable. Use the calculator provided, or this table, to determine the value of c. P(1.22<Z<c)=0.0703 Carry your intermediate computations to at least four decimal places. Round your answer to two decimal places. 0 X $ ?

Answer 1

We are given ,

Z is standard normal variable.

P( 1.22 < z < c ) = 0.0703

0.0703 1.22 с

We have to find c.

P( 1.22 < z < c ) = P( z < c ) - P( z < 1.22 )

P( z < c ) = P( 1.22 < z < c ) + P( z < 1.22 )

P( z < c ) =  0.0703 + P( z < 1.22 )

Using Excel function ,   "=NORMSDIST(Z)"

P( z < 1.22 ) = NORMSDIST(1.22) =0.8888

So, P( z < c ) =  0.0703 + 0.8888 = 0.9591

We have to find c such that area less than c is 0.9591

Using Excel function , "=NORMSINV(probability)"

c = NORMSINV(0.9591) = 1.74

i.e P( 1.22 < z < 1.74 ) = 0.0703