Question

The overhead reach distances of adult females are normally distributed with a mean of 195 cm and a standard deviation of 7.8 cm. a. Find the probability that an individual distance is greater than 207.50 cm. b. Find the probability that the mean for 25 randomly selected distances is greater than 193.20 cm. c. Why can the normal distribution be used in part​ (b), even though the sample size does not exceed​ 30? a. The probability is _______

Answer 1

= 195 cm

= 7.8 cm

a) P(X < A) = P(Z < (A - )/)

P(an individual distance is greater than 207.50 cm) = P(X > 207.5)

= 1 - P(X < 207.5)

= 1 - P(Z < (207.5 - 195)/7.8)

= 1 - P(Z < 1.60)

= 1 - 0.9452

= 0.0548

b) For sampling distribution of mean, P( < A) = P(Z < (A - )/)

Sample size, n = 25

= = 195 cm

=

=

= 1.56

P(mean for 25 randomly selected distances is greater than 193.20 cm) = P( > 193.2)

= 1 - P( < 193.2)

= 1 - P(Z < (193.2 - 195)/1.56)

= 1 - P(Z < -1.15)

= 1 - 0.1251

= 0.8749

c) The population is normally distributed. So, the sampling distribution of mean will also be normally distributed, irrespective of the sample size.