Question

The overhead reach distances of adult females are normally distributed with a mean of

205.5 cm

and a standard deviation of

8.6 cm

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

214.80cm.

b. Find the probability that the mean for

25

randomly selected distances is greater than 204.00 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

standard deviation = = 8.6

P(x > 214.80) = 1 - P(x <214.80 )

= 1 - P[(x - ) / < (214.80- 205.5) / 8.6 ]

= 1 - P(z <1.08 )

Using z table,

= 1 -0.8599

=0.1401

b.

n = 25

= 205.5

= / n = 8.6 / 25 =1.72

P( > 204) = 1 - P( < 204)

= 1 - P[( - ) / < (204-205.5) / 8.6]

= 1 - P(z <-0.17 )

Using z table,    

= 1 - 0.4325

= 0.5625

c.

we can use normal distribution even sample size less than 30 using sampling distribution