Question

The overhead reach distances of adult females are normally distributed with a mean of 195 cm and a standard deviation of 8.3 cm. a. Find the probability that an individual distance is greater than 208.40 cm. b. Find the probability that the mean for 20 randomly selected distances is greater than 192.80 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 nothing. ​(Round to four decimal places as​ needed.)

Answer 1

Solution :

Given that ,

mean = = 195

standard deviation = = 83

(a)

P(x > 208.40) = 1 - P(x < 208.40)

= 1 - P((x - ) / < (208.40 - 195) / 8.3)

= 1 - P(z < 1.6145)

= 1 - 0.9468   

= 0.0532

P(x > ) = 0.0532

Probability = 0.0532

(b)

n = 20

= 195 and

= / n = 8.3 / 20 = 1.8559

P( > 192.80) = 1 - P( < 192.80)

= 1 - P(( - ) / < (192.80 - 195) / 1.8559)

= 1 - P(z < -1.1854)

= 1 - 0.1179

= 0.8821

P( > 192.80) = 0.8821

Probability = 0.8821

(c)

Original distributon is normal .