According to the US Bureau of Labor and Statistics, in 2010, the average America

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According to the US Bureau of Labor and Statistics, in 2010, the average American family spent $124 per week on groceries. Suppose grocery expenditure is normally distributed and suppose the standard deviation was $25. If one randomly selected family's grocery expenditure falls in the bottom 33% of all families, at most what did they spend?

Suppose that there are an average of 8 flaws in 20 yards of fabric. What is the probability that you will find your first flaw within 2 yards?

QUESTION 7 1 points Save Answer According to the US Bureau of Labor and Statistics, in 2010, the average Amenican family spent $124 per week on groceries. Suppose grocery expenditure is normally distributed and suppose the standard deviation was $25. If one randomly selected family's grocery expenditure falls in the bottom 33% of all families, at most what did they spend? a. $113 b.$114.73 c. $115.75 d. $132.25 e, $130.27 f. $135 QUESTION 8 1 points Save Answer Suppose that there are an average of 8 faws in 20 yards of fabric. What is the probability that you will find your first flaw within 2 yards? e" =.0067 e-1.1 -,2865 e- -.6065 e-32 = .0408 e-12-.3012 e-o4 = .6703 e-25= .0821 e-1-.3679 e-025-.7788 =.1353 e-o- .4493 e-a 2-8187 e-0625-5353 e-a 125 e-16-2019 a. ,0067 b..1642 C. .4493 d. .5507 .8358 f. .9933 -.8825 QUESTION 9 1 points Save Answer Suppose x is distributed uniformly on the interval from 35 to 60. Find P(X 50) a. 0 b..4 G. .526 e. .8531 t. Almost 1

Explanation / Answer

7) for 33 percentile ; z =-0.44

hence corresponding value =mean +z*std deviation=124-0.44*25=113

option A is correct

8)

mean distance between 2 flaws =20/8=2.5 yards

hence probability of flaw with in 2 yards =P(X<2)=1-e-2/2.5 =1-e-0.8 =1-0.4493 =0.5507

option D is correct

9)

as probability of a point in a continuous distribution is equal to 0 cause area under a point on a curve is 0,

hence option A is correct

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