3. Christmas Spending. In a recent study done by the National Retail Federation

Question

3. Christmas Spending. In a recent study done by the National Retail Federation found that 2017 Christmas spending for all US households follow a Normal distribution with mean $950 and a standard deviation $60. Use this information to answer the following questions (a) What is the probability that 2017 Christmas spending for a US household is greater than $930? Answer this question by completing parts 3(a)i and 3(a)ii i. Provide the z-score corresponding to the 2017 Christmas spending of $930 ii. Based on your answer in 3(a)i, what is the probability of 2017 Christmas spending for a US household is greater than $930? (b) Free response submission. Provide the z-score corresponding to the 2017 Christmas spending of $1200, and the probability of 2017 Christmas spending for a US household is more than $1200 (c) Find Q1 (First Quartile). Answer this question by completing parts 3(c)i. and 3(c)ii i. Provide the z-score corresponding to Q1 ii. Based on your answer in 3(c)i, provide the value of Q1 2

Explanation / Answer

Solution3ai:

mean=950

sd=60

x=930

z=x-mean/sd

x=930-950/60

=-20/60

=-0.333

z=-0.333

P(Z<-0.333)

Solution3aII

P(Z<-0.333)

rcode is  pnorm(-0.333)

= 0.3695671

ANSWER:0.3695671

Solutionb:

X=1200

Z=1200-950/60

=4.1667

P(Z>4.1667)

1-P(<4.1667)

1-pnomr(-4.1667)

1.545204e-05

ANSWER:

=0.000015

Solutionc:

z score for Q1(25 percentile is -0.674)

-0.674=X-950/60

X=-0.674*60+950

X=909.56

X=910

value of Q1=910

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