Question

According to the last census (2010), the mean number of people per household in the United States is LaTeX: \mu = 2.58 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)

Answer 1

Answer:

P(2.50 < Xbar < 2.66) = 0.046

Step-by-step explanation:

We are given that Population Mean, $\mu$ = 2.58 and Standard deviation, $\sigma$ = 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 = $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ 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 $\leq$ 2.50)

P(Xbar < 2.66) = P( $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{2.66-2.78}{\frac{0.75}{\sqrt{110} } }$ ) = P(Z < -1.68) = 1 - P(Z  1.68)

                                                              = 1 - 0.95352 = 0.04648

P(Xbar $\leq$ 2.50) = P( $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ $\leq$ $\frac{2.50-2.78}{\frac{0.75}{\sqrt{110} } }$  ) = P(Z $\leq$  -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