Question

The
number of hours adults sleep per night is normally distributed with
a mean it 5.5 hours. Assume that the standard deviation is
unknown.

If
75% of adults sleep more than 5.1 hours per night, what is the
variance?

Answer #1

Solution:

The z - distribution of the 75 % is,

P( Z > z ) = 75 %

1 - P( Z < z ) = 0.75

P( Z < ) = 1 - 0.75

P( Z < z ) = 0.25

P( Z < ) = 0.25

z = -0.674

Using z - score formula,

X = z * +

= (X - )/ z

= ( 5.1 - 5.5 ) / -0.674

= 0.5935

^{2}
=
=
0.5935

**The variance is 0.77.**

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