Question

Question Help The overhead reach distances of adult females are normally distributed with a mean of 205.5 cm and standard deviation of 8.9 cm. a. Find the probability that an individual distance is greater than 218.00 cm. b. Find the probability that the mean for 15 randomly selected distances is greater than 203.70 cm. c. Why can the normal distribution be used in part (b), even though the sample size does not exceed 3 a. The probability is .. (Round to four decimal places as needed.)

Answer 1

Solution:

Part a

We are given

µ = 205.5

σ = 8.9

We have to find P(X>218)

P(X>218) = 1 - P(X<218)

Z = (X - µ) / σ

Z = (218 - 205.5) / 8.9

Z = 1.404494

P(Z<1.404494) = P(X<218) = 0.919914

(You can find this probability by using z-table/excel/calculator/etc.)

P(X>218) = 1 - P(X<218)

P(X>218) = 1 - 0.919914

P(X>218) = 0.080086

Required probability = 0.0801

Part b

We are given

µ = 205.5

σ = 8.9

n = 15

We have to find P(x̄ > 203.70)

P(x̄ > 203.70) = 1 - P(x̄ < 203.70)

Z = (X - µ) / [σ/sqrt(n)]

Z = (203.7 - 205.5) / (8.9/sqrt(15))

Z = -0.7833

P(Z<-0.7833) = P(x̄ < 203.70) = 0.216725

(You can find this probability by using z-table/excel/calculator/etc.)

P(x̄ > 203.70) = 1 - P(x̄ < 203.70)

P(x̄ > 203.70) = 1 - 0.216725

P(x̄ > 203.70) = 0.783275

Required probability = 0.7833

Part c

We can use normal distribution in part b because the sampling distribution of sample mean follows normal distribution even though sample size is less than 30.