PRACTICE OF STATISTICS F/AP EXAM
PRACTICE OF STATISTICS F/AP EXAM
6th Edition
ISBN: 9781319113339
Author: Starnes
Publisher: MAC HIGHER
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Chapter 7, Problem R7.3RE

(a)

To determine

To construct: a graph that displays the distribution of birth weights for this population.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given:

  μ=3668σ=511

  PRACTICE OF STATISTICS F/AP EXAM, Chapter 7, Problem R7.3RE , additional homework tip  1

Calculation:

The middle 68% of a normal distribution lies with one standard deviation from the mean of the distribution.

The middle 95% of a normal distribution lies with two standard deviation from the mean of the distribution.

The middle 99.7% of a normal distribution lies with three standard deviation from the mean of the distribution.

Graph:

  PRACTICE OF STATISTICS F/AP EXAM, Chapter 7, Problem R7.3RE , additional homework tip  2

Finding the values that are 1, 2 and 3 standard deviations from the mean

  μ3σ=36683(511)=2135μ2σ=36682(511)=2646μσ=3668511=3157μ+σ=3668+511=4179μ+2σ=3668+2(511)=4690μ+3σ=3668+3(511)=5201

  PRACTICE OF STATISTICS F/AP EXAM, Chapter 7, Problem R7.3RE , additional homework tip  3

(b)

To determine

To construct: a possible graph of the distribution of birth weights for an SRS of size 5 and calculate the range for this sample.

(b)

Expert Solution
Check Mark

Explanation of Solution

Given:

Normal distribution

  μ=3668σ=511

  PRACTICE OF STATISTICS F/AP EXAM, Chapter 7, Problem R7.3RE , additional homework tip  4

Calculation:

A simple random sample of size 5 then contains 5 data values that are roughly centred about a, while all data values are between μ3σ=36683(511)=2135 and μ+3σ=3668+3(511)=5201 .

A possible sample of size 5 is then 2400,3000,3600,3700,4500 .

  Range =max-min=4500-2400=2100

Graph:

Dot plot

  PRACTICE OF STATISTICS F/AP EXAM, Chapter 7, Problem R7.3RE , additional homework tip  5

For each given data value places a dot above the associating number on the number line.

(c)

To determine

To Explain: the value represents.

(c)

Expert Solution
Check Mark

Explanation of Solution

Given:

  PRACTICE OF STATISTICS F/AP EXAM, Chapter 7, Problem R7.3RE , additional homework tip  6

The dots is the dot plot represent the sample range of simple random samples of size n=5 .The dot at 2800 then represents a single simple random sample size n=5 that has a sample ranges of 2800 grams.

(d)

To determine

To Explain: the sample ranges an unbiased estimator of the population range and gives evidence from the graph to support the answer.

(d)

Expert Solution
Check Mark

Explanation of Solution

Given:

  μ=3668σ=511

  PRACTICE OF STATISTICS F/AP EXAM, Chapter 7, Problem R7.3RE , additional homework tip  7

Calculation:

The middle 68% of a normal distribution lies with one standard deviation from the mean of the distribution.

The middle 95% of a normal distribution lies with two standard deviation from the mean of the distribution.

The middle 99.7% of a normal distribution lies with three standard deviation from the mean of the distribution.

  μ3σ=36683(511)=2135

  μ+3σ=3668+3(511)=5201 .

Thus nearly all data values are between 5201 and 2135.

The range is the difference between the maximum and minimum.

  Range =max-min=5201-2135=3066

This then implies that the population range is about 3066 grams.

An estimator is called unbiased, if the centre of the sampling distribution of the estimator is equal to the corresponding population parameter.

In the dot plot, it is observed that all sample ranges are smaller than the population range of 3066 grams. This then implies that the centre of the sampling distribution of the sample ranges is not the population range and thus the sample range is not an unbiased estimator of the population range.

Chapter 7 Solutions

PRACTICE OF STATISTICS F/AP EXAM

Ch. 7.1 - Prob. 11ECh. 7.1 - Prob. 12ECh. 7.1 - Prob. 13ECh. 7.1 - Prob. 14ECh. 7.1 - Prob. 15ECh. 7.1 - Prob. 16ECh. 7.1 - Prob. 17ECh. 7.1 - Prob. 18ECh. 7.1 - Prob. 19ECh. 7.1 - Prob. 20ECh. 7.1 - Prob. 21ECh. 7.1 - Prob. 22ECh. 7.1 - Prob. 23ECh. 7.1 - Prob. 24ECh. 7.1 - Prob. 25ECh. 7.1 - Prob. 26ECh. 7.1 - Prob. 27ECh. 7.1 - Prob. 28ECh. 7.1 - Prob. 29ECh. 7.1 - Prob. 30ECh. 7.1 - Prob. 31ECh. 7.1 - Prob. 32ECh. 7.2 - Prob. 33ECh. 7.2 - Prob. 34ECh. 7.2 - Prob. 35ECh. 7.2 - Prob. 36ECh. 7.2 - Prob. 37ECh. 7.2 - Prob. 38ECh. 7.2 - Prob. 39ECh. 7.2 - Prob. 40ECh. 7.2 - Prob. 41ECh. 7.2 - Prob. 42ECh. 7.2 - Prob. 43ECh. 7.2 - Prob. 44ECh. 7.2 - Prob. 45ECh. 7.2 - Prob. 46ECh. 7.2 - Prob. 47ECh. 7.2 - Prob. 48ECh. 7.2 - Prob. 49ECh. 7.2 - Prob. 50ECh. 7.2 - Prob. 51ECh. 7.2 - Prob. 52ECh. 7.3 - Prob. 53ECh. 7.3 - Prob. 54ECh. 7.3 - Prob. 55ECh. 7.3 - Prob. 56ECh. 7.3 - Prob. 57ECh. 7.3 - Prob. 58ECh. 7.3 - Prob. 59ECh. 7.3 - Prob. 60ECh. 7.3 - Prob. 61ECh. 7.3 - Prob. 62ECh. 7.3 - Prob. 63ECh. 7.3 - Prob. 64ECh. 7.3 - Prob. 65ECh. 7.3 - Prob. 66ECh. 7.3 - Prob. 67ECh. 7.3 - Prob. 68ECh. 7.3 - Prob. 69ECh. 7.3 - Prob. 70ECh. 7.3 - Prob. 71ECh. 7.3 - Prob. 72ECh. 7.3 - Prob. 73ECh. 7.3 - Prob. 74ECh. 7.3 - Prob. 75ECh. 7.3 - Prob. 76ECh. 7.3 - Prob. 77ECh. 7.3 - Prob. 78ECh. 7 - Prob. R7.1RECh. 7 - Prob. R7.2RECh. 7 - Prob. R7.3RECh. 7 - Prob. R7.4RECh. 7 - Prob. R7.5RECh. 7 - Prob. R7.6RECh. 7 - Prob. R7.7RECh. 7 - Prob. T7.1SPTCh. 7 - Prob. T7.2SPTCh. 7 - Prob. T7.3SPTCh. 7 - Prob. T7.4SPTCh. 7 - Prob. T7.5SPTCh. 7 - Prob. T7.6SPTCh. 7 - Prob. T7.7SPTCh. 7 - Prob. T7.8SPTCh. 7 - Prob. T7.9SPTCh. 7 - Prob. T7.10SPTCh. 7 - Prob. T7.11SPTCh. 7 - Prob. T7.12SPTCh. 7 - Prob. T7.13SPTCh. 7 - Prob. AP2.1CPTCh. 7 - Prob. AP2.2CPTCh. 7 - Prob. AP2.3CPTCh. 7 - Prob. AP2.4CPTCh. 7 - Prob. AP2.5CPTCh. 7 - Prob. AP2.6CPTCh. 7 - Prob. AP2.7CPTCh. 7 - Prob. AP2.8CPTCh. 7 - Prob. AP2.9CPTCh. 7 - Prob. AP2.10CPTCh. 7 - Prob. AP2.11CPTCh. 7 - Prob. AP2.12CPTCh. 7 - Prob. AP2.13CPTCh. 7 - Prob. AP2.14CPTCh. 7 - Prob. AP2.15CPTCh. 7 - Prob. AP2.16CPTCh. 7 - Prob. AP2.17CPTCh. 7 - Prob. AP2.18CPTCh. 7 - Prob. AP2.19CPTCh. 7 - Prob. AP2.20CPTCh. 7 - Prob. AP2.21CPTCh. 7 - Prob. AP2.22CPTCh. 7 - Prob. AP2.23CPTCh. 7 - Prob. AP2.24CPTCh. 7 - Prob. AP2.25CPT
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