Question 927260: The average (as measured by the mean) number of children per family in the United States is 0.9 and the standard deviation of the number of children per family in the United States is 1.1. [Source: U.S. Census Bureau, America's Families and Living Arrangements: 2010, Average Number of People per Family Household.] The distribution of the number of children per family in the United States is skewed-right.
What is the probability that a random sample of eighty families in the United States will have a mean of 1.5 children or fewer?
What is the probability that there are 90 or fewer children in a random sample of eighty families?
There is only a 3% chance that a sample of eighty families will have mean number of children greater than what value?
Answer by ewatrrr(24785)   (Show Source):
You can put this solution on YOUR website! Population: mean = .9, sd = 1.1
sample n = 80, SE = 1.1/sqrt(80) = .123
P(xbar ≤ 1.5) = P( z < .6/.123) = P(z < 4.878) = normalcdf(-100, 4.878)= .9999
.......
xbar = 90/80 = 1.125
P(xbar ≤ 1.125) = P(z < .225/.123) = P( z <1.8293) = normalcdf(-100, 1.8293) = .9663
.......
1.1(invNorm(.97) + .9 = x
1.1(1.88) + .9 = 2.968 0r 3 children
........
Using a TI calculator 0r similarly a Casio fx-115 ES plus
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