Household incomes Refer to Exercise 10Here are summary statistics for the state median household incomes:

nMeanSDMinQ1MedQ3Max
5051742.448210.643664146071500095702071836

a. Find and interpret the z-score for North Carolina, with a median household income of $41,553

b. New Jersey had a standardized score of 1.82Find New Jersey’s median household income for that year.

Short Answer

Expert verified

Part (a) North California's median household income is 1.24standard deviations below the mean of all 50states' median household incomes.

Part (b) Median household income in New Jersey is approx. $66,685.80

Step by step solution

01

Part (a) Step 1: Given information

The table is

nMeanSDMinQ1
MedQ3
Max
5051742.448210.643664146071500095702071836

Value, x=41,553

Mean, x=51,742.44

Standard deviation, SD=8210.64

02

Part (a) Step 2: Concept

The formula used:z=(value-mean)(standarddeviation)

03

Part (a) Step 3: Calculation

The z-score is the number of standard deviations a value deviates from the mean.

The value above the mean is indicated by a positive z-score.

Whereas,

The value below the mean is indicated by a negative z-score.

Calculate the z- score:

z=xxSD=41,55351,742.448210.64=10,189.448210.641.24

Therefore,

North California's median household income is 1.24 standard deviations lower than the mean of all 50 states' median household incomes.

04

Part (b) Step 1: Calculation

The z-score is the number of standard deviations a value deviates from the mean.

The value above the mean is indicated by a positive z-score.

Whereas,

A value below the mean is indicated by a negative z-score.

Calculate the z- score:z=xxSD

Substitute the values:

1.82=x51,742.448210.64

Multiply both sides by 8210.64:

14,943.36=x51,742.44

Add 51,742.44 to both sides:

14,943.36+51,742.44=x51,742.44+51,742.44

That becomes

x=66,685.80

Therefore,

For New Jersey, the median household income is approx. $66,685.80

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Where’s the bus? Sally takes the same bus to work every morning. The

amount of time (in minutes) that she has to wait for the bus can be modeled by a uniform distribution on the interval from 0 minutes to 10 minutes.

a. Draw a density curve to model the amount of time that Sally has to wait for the bus. Be sure to include scales on both axes.

b. On what percent of days does Sally wait between 2.5 and 5.3 minutes for the bus?

c. Find the 70th percentile of Sally’s wait times.

Potato chips The weights of 9-ounce bags of a particular brand of potato

chips can be modeled by a Normal distribution with mean μ=9.12ounces and standard deviation σ=0.05ounce. Sketch the Normal density curve. Label the mean and the points that are 1,2, and 3 standard deviations from the mean.

Shoes Refer to Exercise 1. Jackson, who reported owning 22 pairs of shoes, has a standardized score of z=1.10.

a. Interpret this z-score.

b. The standard deviation of the distribution of the number of pairs of shoes owned in this sample of 20 boys is 9.42. Use this information along with Jackson’s z-score to find the mean of the distribution.

The weights of laboratory cockroaches can be modeled with a Normal distribution having mean 80 grams and standard deviation 2 grams. The following figure is the Normal curve for this distribution of weights.

Point C on this Normal curve corresponds to

a.84grams.b.82grams.c.78grams.d.76grams.e.74grams.

Measuring bone density Individuals with low bone density (osteoporosis) have a high risk of broken bones (fractures). Physicians who are concerned about low bone density in patients can refer them for specialized testing. Currently, the most common method for testing bone density is dual-energy X-ray absorptiometry (DEXA). The bone density results for a patient who undergoes a DEXA test usually are reported in grams per square centimeter (g/cm2) and in standardized units. Judy, who is 25years old, has her bone density measured using DEXA. Her results indicate bone density in the hip of 948 g/cm2 and a standardized score of z=1.45The mean bone density in the hip is 956 g/cm2in the reference population of 25-year-old women like Judy.

a. Judy has not taken a statistics class in a few years. Explain to her in simple language what the standardized score reveals about her bone density.

b. Use the information provided to calculate the standard deviation of bone density in the reference population.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free