Step right up! A dot plot of the distribution of height for Mrs. Navard’s class is shown, along with some numerical summaries of the data.

Suppose that Mrs. Navard has the entire class stand on a 6-inch-high platform and then asks the students to measure the distance from the top of their heads to the ground.

a. What shape would this distribution of distance have?

b. Find the mean and median of the distribution of distance.

c. Find the standard deviation and interquartile range (IQR) of the distribution of distance.

Short Answer

Expert verified

Part (a) Right skewed.

Part (b) Mean =67inch

Median =66inch

Part (c) Standard deviation =4.29

Interquartile range =7

Step by step solution

01

Part (a) Step 1: Given information

02

Part (a) Step 2: Explanation

The form of the given distribution is positively skewed or right-skewed, as evidenced by the provided dot plot. The tail of the distribution appears to be on the right side. Furthermore, the mean is greater than the median, indicating the same because the mean is greater than the median when the data is right-skewed.

03

Part (b) Step 1: Explanation

The mean of the distance data is 67inches and the median is 66inches according to the facts provided.

04

Part (c) Step 1: Explanation

According to the data's statistics, the standard deviation is 4.29inchand the interquartile range is the difference between the data's third and first quartiles. The first quartile score is 63while the third quartile score is 70

Therefore, the interquartile range =7063=7

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

IQ test scores Scores on the Wechsler Adult Intelligence Scale (an IQ test) for the 20- to 34-year-old age group are approximately Normally distributed with μ=110 and σ=25 .

a. What percent of people aged 20 to 34 have IQs between 125 and 150?

b. MENSA is an elite organization that admits as members people who score in the top 2% on IQ tests. What score on the Wechsler Adult Intelligence Scale would an individual aged 20 to 34 have to earn to qualify for MENSA membership?

Used cars, cheap! Refer to Exercise 24Suppose each car’s price is reduced by 10% instead of $500What effect will this discount have on each of the following characteristics of the resulting price distribution?

a. Shape

b. Median

c. Interquartile range (IQR)

Deciles The deciles of any distribution are the values at the 10th,20th,,and 90th percentiles. The first and last deciles are the 10th and the 90th percentiles, respectively. What are the first and last deciles of the standard Normal distribution?

SAT versus ACT Eleanor scores 680 on the SAT Mathematics test. The distribution of SAT Math scores is symmetric and single-peaked with mean 500 and a standard deviation 100 Gerald takes the American College Testing (ACT) Mathematics test and scores 29ACT scores also follow asymmetric, single-peaked distribution—but with mean 21 and standard deviation of 5 Find the standardized scores for both students. Assuming that both tests measure the same kind of ability, who has the higher score?

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.

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