SOLUTION: A local survey found that families of 5 spend, on average $167.45 in groceries a week. Assume the standard deviation is $23.78. Find the percentage of families that spend over $190

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Question 1056716: A local survey found that families of 5 spend, on average $167.45 in groceries a week. Assume the standard deviation is $23.78. Find the percentage of families that spend over $190.00 a week on groceries. Assume the variable is normally distributed. Choose the correct value:

1.) 7.11%
2.) 1.17%
3.) 17.11%
4.) 11.17%
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
z=(190-167.45)/23.78=0.948
probability z>0.948 is 0.1715
3 is closest.