Question

The mean height of an adult giraffe is 18 feet. Suppose that the distribution is normally distributed with standard deviation 0.8 feet. Let X be the height of a randomly selected adult giraffe. Round all answers to 4 decimal places where possible.

Answer 1

Let X be the height of a randomly selected giraffe

X~ N( 18, 0.82)

c) z- score of a giraffe that is 19 foot tall

z = $\frac{x-\mu }{\sigma }$ = $\frac{19-18}{0.8 }$ = 1.25

d) P( X < 17.7) = P( $\frac{x-\mu }{\sigma }$ < $\frac{17.7-18}{0.8 }$ )

= P( z < -0.37)

= 1- P(z < 0.37)

= 1- 0.64431

= 0.3557

e) P( 17.7 < X < 18.4) = P( $\frac{17.7-18}{0.8 }$ < $\frac{x-\mu }{\sigma }$ < $\frac{18.4-18}{0.8 }$ )

= P( -0.37< z < 0.5)

= P( z < 0.5 ) - P( z < -0.37)

= P( z < 0.5) -1 + P( z < 0.37)

= 0.69146 - 1 + 0.64431

= 0.33577

f) For the 70th percentile

P( Z < z) =0.7

P( Z < 0.524 ) = 0.7 { from the percentile table)

z = 0.524

  $\frac{x-\mu }{\sigma }$ = 0.524

$\frac{x-18}{0.8 }$ = 0.524

x = 18 + (0.524 *0.8)

x = 18.4192

So, 18. 4192 is the height of the giraffe corresponding to the 70th percentile