/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q. 7 T2.7. If the heights of American... [FREE SOLUTION] | 91影视

91影视

T2.7. If the heights of American men follow a Normal distribution, and 99.7% have heights between 5'0'' and 7'0'', what is your estimate of the standard deviation of the height of American men?
(a) 1''
(b) 3''
(c) 4''
(d) 6''
(e) 12''

Short Answer

Expert verified

The standard deviation of the height of American men is option (c)4''.

Step by step solution

01

Given information

The heights of American men follow a Normal distribution, and 99.7%have heights between 5'0'' and 7'0''.

02

Explanation

According to the 68-95-99.7rule, note that 99.7%of the observation lies within three standard deviations of the mean (3)

Then,
3=5(1)

+3=7(2)

Subtract the equations (1) and (2), to obtain the value of standard deviation as,(+3)(3)=756=2=26=0.3333

It is known that 12inches 12*constitutes 1 foot 1'

Hence, the standard deviation is obtained as

=0.333312"=4"

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 91影视!

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

The distribution of heights of adult American men is approximately Normal with mean 69inches and standard deviation of 2.5inches. Draw a Normal curve on which this mean and standard deviation are correctly located. (Hint: Draw the curve first, locate the points where the curvature changes, then mark the horizontal axis.)

Use Table A to 铿乶d the proportion of observations from the standard Normal distribution that satis铿乪s each of the following statements. In each case, sketch a standard Normal curve and shade the area under the curve that is the answer to the question. Use your calculator or the Normal Curve applet to check your answers.

More Table A practice

(a) zis between 鈭1.33and 1.65

(b) zis between 0.50and1.79

The weights of laboratory cockroaches follow a Normal distribution with mean 80grams and standard deviation 2grams. The figure below is the Normal curve for this distribution of weights.

Point C on this Normal curve corresponds to

(a) 84grams.

(c) 78grams.

(e) 74grams

(b) 82grams.

(d) 76grams.

T2.4. The figure shows a cumulative relative frequency graph of the number of ounces of alcohol consumed per week in a sample of 150adults. About what percent of these adults consume between 4and 8ounces per week?

(a)20%(b)40%(c)50%(d)60%(e)80%

Normal is only approximate: ACT scores Scores on the ACT test for the 2007 high school graduating class had mean 21.2and standard deviation 5.0. In all, 1,300,599students in this class took the test. Of these, 149,164had scores higher than 27and another 50,310had scores exactly 27. ACT scores are always whole numbers. The exactly Normal N(21.2,5.0)distribution can include any value, not just whole numbers. What鈥檚 more, there is no area exactly above 27under the smooth Normal curve. So ACT scores can be only approximately Normal. To illustrate this fact, find

(a) the percent of 2007ACT scores greater than 27.

(b) the percent of 2007ACT scores greater than or equal to 27.

(c) the percent of observations from theN(21.2,5.0) distribution that are greater than 27. (The percent greater than or equal to 27 is the same, because there is no area exactly over 27.)

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.