Assume a standard deviation of LaTeX: \sigma = 0.75σ = 0.75. You plan to take a random sample of 110 households, what is the probability the sample mean household size is between 2.5 and 2.66 people? (round your probabilities to three decimal places)

P(2.50 < Xbar < 2.66) = 0.046

Step-by-step explanation:

We are given that Population Mean, = 2.58 and Standard deviation, = 0.75

Also, a random sample (n) of 110 households is taken.

Let Xbar = sample mean household size

The z score probability distribution for sample mean is give by;

             Z = ~ N(0,1)

So, probability that the sample mean household size is between 2.50 and 2.66 people = P(2.50 < Xbar < 2.66)

P(2.50 < Xbar < 2.66) = P(Xbar < 2.66) - P(Xbar 2.50)

P(Xbar < 2.66) = P( < ) = P(Z < -1.68) = 1 - P(Z 1.68)

                                                              = 1 - 0.95352 = 0.04648

P(Xbar 2.50) = P( ) = P(Z -3.92) = 1 - P(Z < 3.92)

                                                              = 1 - 0.99996 = 0.00004

Therefore, P(2.50 < Xbar < 2.66) = 0.04648 - 0.00004 = 0.046

x= 149. 24+7=31. 180-31=149

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