A recent study done by the National Retail Federation found that 2019 back-to-school spending for all US households who have school-aged children follows a Normal distribution with mean $697 and a standard deviation $120. Use this information to answer the following questions. (a) What is the probability that 2019 back-to-school spending for a US household with school- aged children is greater than $893? Answer this question by completing parts 2(a)i and 2(a)ii. i. Provide the z-score corresponding to the 2019 back-to-school spending of $893. ii. Based on your answer in 2(a)i, what is the probability of 2019 back-to-school spending for a US household with school-aged children is greater than $893? (b) Free response submission. Provide the 2-score corresponding to the 2019 back-to- school spending of $1,200, and the probability of 2019 back-to-school spending for a house- hold with school-aged children is less than $1,200. (c) Find Q3 (Third Quartile). Answer this question by completing parts 2(c)i. and 2(c)ii. i. Provide the z-score corresponding to Q3. ii. Based on your answer in 2(c)i, provide the value of Q3. (d) Find Q1 (First Quartile). Answer this question by completing parts 2(d)i and 2(d)ii. i. Provide the z-score corresponding to Q1. ii. Based on your answer in 2(d)i, provide the value of Qı. (e) What is the value of the IQR for the distribution of 2019 back-to-school spending for a US household with school-aged children? (f) Free response submission. Interpret the value of the IQR from question 2e within the context of the problem. (g) What is the proportion of 2019 back-to-school spending within 1.50 standard deviations of the mean? Answer this question by completing parts 2(g)i. through 2(g)iii. i. Provide the SMALLER 2-score corresponding to the above statement. ii. Provide the LARGER z-score corresponding to the above statement. iii. Based on your answers in 2(g)i and 2(g)ii, what proportion of 2019 back-to-school spending is within 1.50 standard deviations of the mean? (h) What is the 2019 back-to-school spending amount such that only 3% of households with school-age children spend more than this amount? i. Provide the z-score corresponding to the top 3%. ii. Based on your answer in 2(h)i, provide the spending amount such that only 3% of households with school-age children spend more than this amount? (i) Free response submission. Which US household is more unusual, a US household with back-to-school spending of $600 or a US household with back-to-school spending of $900? Explain your answer. (j) Free response submission. Let's say the Smith family spent $815 on buying school supplies this fall. Provide an interpretation of the Smith family's 2019 back-to-school spending, i. e. what can you say about the percentage of all other US households with school-age children that have a higher back-to-school spending than the Smith family?

Step-by-step explanation:

Hello!

The working variable is:

X: Back-to-school expense of a US household with school-aged children.

X~N(μ;σ²)

μ= $697

σ= $120

a. What is the probability that 2019 back-to-school spending for a US household with school-aged children is greater than $893?

Symbolically: P(X>$893)

First, you standardize the probability using Z= (X-μ)/σ ~N(0;1)

P(X>$893)= P(Z>(893-697)/120)= P(Z>1.63)

To resolve this question you have to use the table of cumulative probabilities for the standard normal distribution. These tables accumulate probabilities from the left, symbolically P(Z≤Z₀), so to reach probabilities greater than a Z₀ value you have to subtract the cumulative probability until that value from the maximum probability value 1:

P(Z>1.63)= 1 - P(Z≤1.63)= 1 - 0.94845= 0.05155

b. Provide the Z-score corresponding to the 2019 back-to-school spending of $1,200, and the probability of 2019 back-to-school spending for a household with school-aged children is less than $1,200.

P(X<$1200) = P(Z<(1200-697)/120)= P(Z<4.19)= 1

According to the empirical rule of the normal distribution, 99% of the data is between μ ± 3σ. This, logically, applies to the standard normal distribution. Considering that the distribution's mean is zero and the standard deviation is one, then 99% of the probabilities under the standard normal distribution are within the Z values: -3 and 3, values below -3 will have a probability equal to zero and values above 3 will have probability equal to one.

c. Find Q3 (Third Quartile).

Q3 in the value that marks three-quarters of the distribution, in other words, it has 75% of the distribution below it and 25% above, symbolically:

P(Z≤c)=0.75

In this case, you have to look in the center of the right Z-table (positive) for the probability of 0.75 and then the margins to find the Z-score that belongs to that cumulative probability:

c= 0.674

Now you reverse the standardization to see what value of X belongs to the Q3:

c= (X-μ)/σ

X= (c*σ)+μ

X= (0.674*120)+697= $777.88

d. Find Q1 (First Quartile)

To resolve this you have to follow the same steps as in c., just that this time you'll look for the value that marks the first quarter of the distribution, symbolically:

P(Z≤d)= 0.25

In this case, since the probability is below 0.5 you have to look for the Z value in the left table (negative).

d= -0.674

d= (X-μ)/σ

X= (d*σ)+μ

X= (-0.674*120)+697= $616.12

e. What is the value of the IQR for the distribution of 2019 back-to-school spending for a US household with school-aged children?

IQR= Q3-Q1= $777.88 - $616.12= $161.76

f. Interpret the value of the IQR from question 2e within the context of the problem.

$161.76 represents the distance between 75% of the Back-to-school expense of a US household 25% of the Back-to-school expense of US households.

g. What is the proportion of 2019 back-to-school spending within 1.50 standard deviations of the mean?

"Within 1.50 standard deviations of the mean" can be symbolized as "μ ± 1.5σ" or "μ - 1.5σ≤ Z ≤μ + 1.5σ"

P(μ - 1.5σ≤ Z ≤μ + 1.5σ)

Since the mean is zero and the standard deviation is one:

P(-1.5 ≤ Z ≤ 1.5)= P(Z≤1.5) - P(Z≤-1.5)= 0.933 - 0.067= 0.866

h. What is the 2019 back-to-school spending amount such that only 3% of households with school-age children spend more than this amount?

The "top" 3% means that you are looking for a value of the variable that has above it 0.03 of probability and below it 0.97%, first you look for this value under the standard normal distribution and then you reverse the standardization to reach the corresponding value of the variable:

P(Z>h)= 0.03 ⇒ P(Z≤h)=0.97

h= 1.881

h= (X-μ)/σ

X= (h*σ)+μ

X= ( 1.881*120)+697= $922.72

i. Which US household is more unusual, a US household with back-to-school spending of $600 or a US household with back-to-school spending of $900?

Under this kind of distribution, the "most usual" values are around the center (near the mean) and the "unusual" values will find themselves in the tails of the Gaussian bell.

To check which one is more unusual you have to see their distance with respect to the mean.

(X-μ)/σ

(600-697)/120= -0.8083

(900-697)/120= 1.69

An expense of $900 is more unusual than an expense of $600 (600 is almost the expected expenses)

j. Let's say the Smith family spent $815 on buying school supplies this fall. Provide an interpretation of the Smith family's 2019 back-to-school spending, i.e. what can you say about the percentage of all other US households with school-age children that have higher back-to-school spending than the Smith family?

P(X>$815) = P(Z>(815-697)/120)= P(Z>0.98)

1-P(Z≤0.983)= 0.837

83.7% of the families will have back-to-school expenses of $815 or more.

I hope it helps!

12.5 feet

step-by-step explanation:

answer: (c) m∠qpo + (2x + 16)° = 180°

step-by-step explanation:

the opposite angles of a quadrilateral are supplementary.

so, ∠o + ∠q = 180°   and   ∠p + ∠r = 180°

since ∠r = 2x + 16°, we can use substitution as follows:

∠p + ∠r = 180°

∠p + (2x + 16)° = 180°