Question

6.4.6-T on Hep The overhead reach distances of adult females are normally distributed with a mean of 2025 cm and a standard deviation of 86 a. Find the probability that an individual distance is greater than 212 50 cm b. Find the probability that the mean for 15 randomly selected distances is greater than 201.20 cm the sample size does not not exceed 30

Answer 1

Solution :

Given that mean μ = 202.5 , standard deviation σ = 8.6

a. => P(x > 212.50) = P((x - μ)/σ > (212.50 - 202.5)/8.6)

= P(Z > 1.1628)

= 1 − P(Z < 1.1628)

= 1 − 0.8770

= 0.1230

b. when n = 15

=> P(x > 201.20) = P((x - μ)/(σ/sqrt(n)) > (201.20 - 202.5)/(8.6/sqrt(15)))

= P(Z > -0.5855)

= P(Z < 0.5855)

= 0.7224

c. => The normal distribution can be used because the original population has a normal distribution

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Solution :

Given that mean μ = 76.0 , standard deviation σ = 12.5

a. => P(69 < x < 83) = P((69 - 76)/12.5 < (x - μ)/σ < (83 - 76)/12.5)

= P(-0.56 < Z < 0.56)

= 0.4246