Question

The random variable x has a normal distribution with mean 23 and standard deviation 5.

Find the value d such that P(20<X<d)=0.4641

Answer #1

Solution

P( 20 < x < d ) = 0.4641

P[(20 - 23 )/ 5 ) < (x - ) / < d) ] = 0.4641

= P(-0.60 < z < d) = 0.4641

= P(z < d) - P(z < -0.60) = 0.4641

Using z table,

= P(z < d) - 0.2743 = 0.4641

= P(z < d) = 0.4641 + 0.2743

= P(z < d) = 0.7384

= P(z < 0.64) = 0.7384

z = 0.64

Using z-score formula,

d = z * +

d = 0.64 * 5 + 23

d = 26.2

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