Question

2. [18] According to the Statistical Bureau’s consumer expenditure survey in 2017 for the island of Maui, the average household spent $1000 annually on gasoline. Suppose that the annual household expenditure on gasoline is normally distributed. Since some households spend nothing on gasoline, let us assume that the range of annual household gasoline expenditures varies from zero to $2500, so that the standard deviation is approximately $500.

a. [3] Does the survey represent cross-sectional or time series data? Explain.

b. [3]What is the probability that a household spends more than $1600 per year on gasoline? Standard deviation= $500 Average = $1000 Z=(X-

c. [3] Ninety five percent of all gasoline expenditures would be expected to fall within what symmetrical limits?

d. [3] What is the probability that a household chosen at random spent between $800 and $1500 on gasoline annually?

e. [3] A new survey indicated that 10% of households now spend more than $1600 per year on gasoline, what is the new mean annual expenditure on gasoline?

f. [3]What is the probability that a random sample of 50 households spent less than an average of $750 per year on gasoline?

Answer 1

sollution: from the given the data, (A) Yes, the survey represents cross sectional data, since data is collected on number of households over a period    (B) X: household spend on gasoline per year

X~N(mu=1000 , sd=500)

Z=(X-mu)/sd ~N(0,1)

Probability that household spend more than $1600 = P(X>1600)

P(X>1600) = P((X-mu)/sd > (1600-mu)/sd) = P((X-mu)/sd > (1600-1000)/500)

P(X>1600) = P(Z > 1.2) = 0.1150697

hence, the Probability =, 0.1150697

(C)

95% CI is (x_bar - sd*z_ , x_bar + sd*z_ )

95% CI is (1000 - 500* 1.960 ,  1000 + 500* 1.960) i.e. (20, 1980)

hence 95% of gasoline expenditures would be expected to fall within (20, 1980)

(D)

Probability that household spents on gasoline in between 800 and 1500 =  P(800 < X < 1500)

P(800 < X < 1500) = P( (800-mu)/sd < (X-mu)/sd < (1500-mu)/sd)

P(800 < X < 1500) = P( (800-1000)/500 < Z < (1500-1000)/500)

P(800 < X < 1500) = P( -0.4 < Z < 1) = 0.4967665

Probability that household spents on gasoline in between 800 and 1500 =  P(800 < X < 1500) = 0.4967665

hence the Probability = 0.4967665

(E)

P(X>1600) =0.10

P(X>1600) = P((X-mu)/sd > (1600-mu)/sd) = P(Z > (1600-mu)/500) = 0.10

hence (1600-mu)/500 = 1.281552  

mu = 1600 - 500* 1.281552 = 959.224

hence mean annual expenditure on gasoline will be $959.224

(F) probabilty that sample of 50 households spent less than 750 = P(x_bar < 750)

P(x_bar < 750) = P( (x_bar- mu)*sqrt(50)/sd < (750-mu)*sqrt(50)/sd)

P(x_bar < 750) = P(Z< (750-1000)*sqrt(50)/500) = P(Z < -3.535534) = 0.000203476

P(x_bar < 750) = 0.000203476

Hence , the  probabilty = 0.000203476

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