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)i 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 S930? (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. (d) Find Qs (Third Quartile). Answer this question by completing parts 3(d)i and 3(d)ii. i. Provide the z-score corresponding to Qs ii. Based on your answer in 3(d)i, provide the value of Qs households? context of the problem (e) What is the value of the IQR for the distribution of 2017 Christmas spending for all US (f) Free response submission. Interpret the value of the IQR from question 3e within the

Answer 1

Let X denote the 2017 Christmas spending random variable. Now we are given that :

$X \sim N(\mu = 950, \sigma^2=60^2)$

(a)(i)

The z-score corresponding to the 2017 christmas spending of $930 is given by :

$z=\frac{ 930-\mu}{ \sigma}=\frac{930-950}{60}=\frac{-20}{60}=-0.3333$

(ii)

The probability of 2017 christmas spending for a US househols is greater than $930 is given by :

P(X>930) = P(Z > -0.3333) where Z is the standard normal variable, Z~N(0,1)

Thus, P(X>930) = P(Z>-0.3333) = 0.63056

(b)

The z-score corresponding to the 2017 christmas spending of $1200 is given by :

$z=\frac{ 1200-\mu}{ \sigma}=\frac{1200-950}{60}=\frac{250}{60}=4.1667$

The probability of 2017 christmas spending for a US househols is greater than $1200 is given by :

P(X>1200) = P(Z > 4.1667) where Z is the standard normal variable, Z~N(0,1)

Thus, P(X>1200) = P(Z>4.1667) = 0.000015

(c)(i)

The z-score corresponding to Q₁ can be found from the tables of standard normal distributions as:

z = -0.67449

(ii)

Now to find the Q₁ we equate the z value obtained in Part (c)(i) with the standardized value of  Q₁ :

$\frac{ Q_1-\mu}{ \sigma}=-0.67449$

$=>\frac{ Q_1-950}{ 60}=-0.67449$

$=>Q_1=909.53$

(d)(i)

The z-score corresponding to Q₃ can be found from the tables of standard normal distributions as:

z = 0.67449

(ii)

Now to find the Q₃ we equate the z value obtained in Part (d)(i) with the standardized value of  Q₃ :

$\frac{ Q_3-\mu}{ \sigma}=0.67449$

$=>\frac{ Q_3-950}{ 60}=0.67449$

$=>Q_3=990.47$

(e)

IQR = Q₃ - Q₁ = 990.47-909.53 = 80.94

(f)

The IQR of 80.94 means that the middle 50% of the spending of US households for 2017 christmas has a range of $80.94