The overhead reach distances of adult females are normally distributed with a mean of 195 cm and a standard deviation of 8.9 cm. a. Find the probability that an individual distance is greater than 207.50 cm. b. Find the probability that the mean for 20 randomly selected distances is greater than 193.50 cm. c. Why can the normal distribution be used in part​ (b), even though the sample size does not exceed​ 30

a) P(X > 207.50) = 0.0808

b) P(X-bar > 193.5) = 0.7734

c) It matters only in sampling distribution when we does not know the shape of distribution.

Step-by-step explanation:

Given - The overhead reach distances of adult females are normally distributed with a mean of 195 cm and a standard deviation of 8.9 cm.

To find - a. Find the probability that an individual distance is greater than 207.50 cm.

              b. Find the probability that the mean for 20 randomly selected distances is greater than 193.50 cm.

              c. Why can the normal distribution be used in part​ (b), even though the sample size does not exceed​ 30 ?

Proof -

Given that,

μ = 195 cm

σ = 8.9 cm

a)

P(X > 207.50) = 1 - P(X < 207.5)

                       = 1 - P(Z < )

                       = 1 - P(Z < 1.4)

                       = 1 - 0.9192

                       = 0.0808

⇒P(X > 207.50) = 0.0808

b)

n = 20

μₓ- = μ = 195

σₓ⁻ = = = 1.99

Now,

P(X-bar > 193.5) = 1 - P(X-bar < 193.5)

                          = 1 - P(Z < )

                          = 1 - P(Z < -0.75)

                          = 1 - 0.2266

                          = 0.7734

⇒P(X-bar > 193.5) = 0.7734

c)

Given that If the population is normally distributed , then the sample does not have to be greater than 30.

It matters only in sampling distribution when we does not know the shape of distribution.

a28=58

step-by-step explanation:

a1=a1+(n-1)d

a28= 4+ (28-1)2

a28=4+(27)(2)

a28=4+54

a28=58