Question

The overhead reach distances of adult females are normally distributed with a mean of 202.5 cm and a standard deviation of 8.9 cm.

a. Find the probability that an individual distance is greater than 211.80 cm.

b. Find the probability that the mean for 20 randomly selected distances is greater than 200.70 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 = = 202.5

standard deviation = = 8.9

P(x >211.80 ) = 1 - P(x <211.80 )

= 1 - P[(x - ) / < (211.80 - 202.5) /8.9 ]

= 1 - P(z <1.05 )

Using z table,

= 1 -0.8531

=0.1469

(b)n = 20

= 202.5

= / n = 8.9 / 20 = 1.9901

P( >200.70 ) = 1 - P( <200.70 )

= 1 - P[( - ) / < (200.70 - 202.5) /1.9901 ]

= 1 - P(z <-0.90 )

Using z table,    

= 1 - 0.1841

= 0.8159

(c)normal distribution can use for any sample size