Question

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

205.5 cm205.5 cm

and a standard deviation of

7.8 cm7.8 cm.

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

215.50215.50

cm.b. Find the probability that the mean for

2525

randomly selected distances is greater than 203.70 cm.203.70 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 = = 205.5

standard deviation = = 7.8

(a)

P(x > 215.50) = 1 - P(x < 215.50)

= 1 - P((x - ) / < (215.50 - 205.5) / 7.8)

= 1 - P(z < 1.28)

= 1 - 0.8997   

= 0.1003

Probability = 0.1003

(b)

n = 25

= 205.5 and

= / n = 7.8 / 25 = 7.8 / 5 = 1.56

P( > 203.70) = 1 - P( < 203.70)

= 1 - P(( - ) / < (203.70 - 205.5) / 1.56)

= 1 - P(z < -1.15)

= 1 - 0.1251

= 0.8749

Probability = 0.8749

(c)

Original distribution is normal .