Question

The overhead reach distances of adult females are normally distributed with a mean of 205.5 cm and a standard deviation of 8 cm. a. Find the probability that an individual distance is greater than 218.00 cm. b. Find the probability that the mean for 15 randomly selected distances is greater than 203.70 cm. c. Why can the normal distribution be used in part? (b), even though the sample size does not exceed? 30?

Answer 1

Solution:

Given that,

mean = = 205.5 cm.

standard deviation = = 8 cm.

a ) p ( x > 218..00 )

= 1 - p (x < 218.00 )

= 1 - p ( x -  / ) < ( 218.00 - 205.5 / 8)

= 1 - p ( z < 12.5 /8 )

= 1 - p ( z < 1.56 )

Using z table

= 1 - 0.9406

= 0.0594

Probability = 0.0594

b ) n = 15

So,

   = 205.5 cm

=  ( /n) = (8 / 15 ) = 2.0656 cm

p (> 203..70 )

= 1 - p ( < 203.70 )

= 1 - p ( -  / ) < ( 203..70 - 205.5 / 2.0656 )

= 1 - p ( z < -1.8 /2.0656 )

= 1 - p ( z < - 0.87 )

Using z table

= 1 - 0.1922

= 0.8078

Probability = 0.8078

c ) Given that distribution approctimatly normal.