Question

Let us assume that from a population with mean ?=100 and standard deviation ?=15 a sample random variable of ?=900 is selected.
a. What is the probability ?(?̅<101.1)?
b. What is the probability ?(?̅>101.5)?
c. What is the probability ?(99.3<?̅<100.5)?

Answer 1

Let X be the random variable that is from a population with mean = 100 and standard deviation = 15.

A sample of n = 900 is selected.

According to the central limit theorem, if the population distribution has mean and variance ², then the sample mean will approximately be normally distributed with mean and variance ²/n, where n is the sample size, regardless of the population distribution.

Using the central limit theorem, Normal ( = 100, ²/n = 15²/900)

a. Answer :

P( < 101.1) = P(Z < (101.1 - 100) / (15/))

                       = P(Z < 2.2)

                       = 0.98610       ...........(value from standard normal table)

Therefore, the probability P( < 101.1) is 0.9861

b. Answer :

P( > 101.5) = P(Z > (101.5 - 100) / (15/))

                       = P(Z > 3)

                       = 1 - P(Z < 3)

                       = 1 - 0.99865     ...........(value from standard normal table)

                       = 0.00135

Therefore, the probability P( > 101.5) is 0.0014

c. Answer :

P(99.3 < < 100.5) = P((99.3 - 100) / (15/) < Z < (100.5 - 100) / (15/))

                                  = P(-1.4 < Z < 1)

                                  = P(Z < 1) - P(Z < -1.4)

                                  = P(Z < 1) - P(Z > 1.4)    ............( P(Z < -a) = P(Z > a))

                                  = P(Z < 1) - (1 - P(Z < 1.4))

                                  = 0.84134 - (1 - 0.91924)     ...........(values from standard normal table)

                                  = 0.76058

Therefore, the probability P(99.3 < < 100.5) is 0.7606