/*! 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} Problem 7 Find the population variance and... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find the population variance and standard deviation or the sample variance and standard deviation as indicated. $$ \text { Population: } 3,6,10,12,14 $$

Short Answer

Expert verified
Variance = 16, Standard Deviation = 4

Step by step solution

01

Calculate the Mean of the Population

First, find the mean (average) of the population values. The values are 3, 6, 10, 12, and 14. The mean is calculated by the formula: \[ \text{Mean} = \frac{\text{Sum of all values}}{\text{Number of values}} \ \text{Mean} = \frac{3 + 6 + 10 + 12 + 14}{5} = \frac{45}{5} = 9 \]
02

Calculate Each Deviation from the Mean

Subtract the mean from each value to find the deviation of each value: \[ 3 - 9 = -6, \ 6 - 9 = -3, \ 10 - 9 = 1, \ 12 - 9 = 3, \ 14 - 9 = 5 \]
03

Square Each Deviation

Square each of the deviations: \[ (-6)^2 = 36, \ (-3)^2 = 9, \ 1^2 = 1, \ 3^2 = 9, \ 5^2 = 25 \]
04

Calculate the Mean of the Squared Deviations

Find the mean of these squared deviations to get the population variance: \[ \text{Variance} = \frac{36 + 9 + 1 + 9 + 25}{5} = \frac{80}{5} = 16 \]
05

Calculate the Population Standard Deviation

To find the population standard deviation, take the square root of the variance: \[ \text{Standard Deviation} = \root 2 \/16 = 4 \]

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 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

calculate mean
To start any variance or standard deviation calculations, you first need to find the mean (average) of your dataset.
Simply, the mean is the sum of all values divided by the number of values.
The formula to calculate the mean is: \[ \text{Mean} = \frac{\text{Sum of all values}}{\text{Number of values}} \]
For example, consider the population values: 3, 6, 10, 12, and 14.
First, sum them up: \[ 3 + 6 + 10 + 12 + 14 = 45 \]
Then, divide by the number of values (which is 5): \[ \text{Mean} = \frac{45}{5} = 9 \]
So, the mean of these values is 9.
deviation from mean
Next, we move on to finding how each value differs from the mean.
These differences are called deviations.
To find the deviation of each value, subtract the mean from each value in the dataset.
Let's go through it with our dataset:
1. Calculate for 3: \[ 3 - 9 = -6 \]
2. For 6: \[ 6 - 9 = -3 \]
3. For 10: \[ 10 - 9 = 1 \]
4. For 12: \[ 12 - 9 = 3 \]
5. For 14: \[ 14 - 9 = 5 \]
So, the deviations are -6, -3, 1, 3, and 5 respectively.
These deviations tell us how far each value lies from the mean.
squared deviations
After finding the deviations, the next step is to square each of these deviations.
Why square them? To ensure all values are positive and give more weight to larger differences.
Squaring a number means multiplying it by itself.
Let's square our deviations:
1. For -6: \[ (-6)^2 = 36 \]
2. For -3: \[ (-3)^2 = 9 \]
3. For 1: \[ 1^2 = 1 \]
4. For 3: \[ 3^2 = 9 \]
5. For 5: \[ 5^2 = 25 \]
So, the squared deviations are 36, 9, 1, 9, and 25 respectively.
mean of squared deviations
Now, we need to find the average of these squared deviations.
This average is known as the variance.
Use the same process as finding the mean but with the squared deviations instead.
Sum up the squared deviations: \[ 36 + 9 + 1 + 9 + 25 = 80 \]
Then, divide by the number of values: \[ \text{Variance} = \frac{80}{5} = 16 \]
So, the variance of this population is 16.
population standard deviation
Finally, to find the population standard deviation, take the square root of the variance.
This step reverses the squaring we did earlier and returns the value to the original unit.
Use the square root function: \[ \text{Standard Deviation} = \root 2 \/16 = 4 \]
Therefore, the population standard deviation is 4.
This value tells us how spread out the values are in the population.

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 average 20 - to 29 -year-old man is 69.6 inches tall, with a standard deviation of 3.0 inches, while the average 20 - to 29 -year-old woman is 64.1 inches tall, with a standard deviation of 3.8 inches. Who is relatively taller, a 75 -inch man or a 70 -inch woman?

Babies born after a gestation period of 32-35 weeks have a mean weight of 2600 grams and a standard deviation of 660 grams. Babies born after a gestation period of 40 weeks have a mean weight of 3500 grams and a standard deviation of 470 grams. Suppose a 34 -week gestation period baby weighs 3000 grams and a 40 -week gestation period baby weighs 3900 grams. What is the \(z\) -score for the 34 -week gestation period baby? What is the \(z\) -score for the 40 -week gestation period baby? Which baby weighs less relative to the gestation period?

In December \(2014,\) the average price of regular unleaded gasoline excluding taxes in the United States was \(\$ 3.06\) per gallon, according to the Energy Information Administration. Assume that the standard deviation price per gallon is \(\$ 0.06\) per gallon to answer the following. (a) What minimum percentage of gasoline stations had prices within 3 standard deviations of the mean? (b) What minimum percentage of gasoline stations had prices within 2.5 standard deviations of the mean? What are the gasoline prices that are within 2.5 standard deviations of the mean? (c) What is the minimum percentage of gasoline stations that had prices between \(\$ 2.94\) and \(\$ 3.18 ?\)

The following data represent the pulse rates (beats per minute) of nine students enrolled in a section of Sullivan's Introductory Statistics course. Treat the nine students as a population. $$ \begin{array}{lc} \text { Student } & \text { Pulse } \\ \hline \text { Perpectual Bempah } & 76 \\ \hline \text { Megan Brooks } & 60 \\ \hline \text { Jeff Honeycutt } & 60 \\ \hline \text { Clarice Jefferson } & 81 \\ \hline \text { Crystal Kurtenbach } & 72 \\ \hline \text { Janette Lantka } & 80 \\ \hline \text { Kevin McCarthy } & 80 \\ \hline \text { Tammy Ohm } & 68 \\ \hline \text { Kathy Wojdyla } & 73 \\ \hline \end{array} $$ (a) Determine the population mean pulse. (b) Find three simple random samples of size 3 and determine the sample mean pulse of each sample. (c) Which samples result in a sample mean that overestimates the population mean? Which samples result in a sample mean that underestimates the population mean? Do any samples lead to a sample mean that equals the population mean?

The acidity or alkalinity of a solution is measured using pH. A pH less than 7 is acidic; a pH greater than 7 is alkaline. The following data represent the \(\mathrm{pH}\) in samples of bottled water and tap water. $$ \begin{array}{lllllll} \hline \text { Tap } & 7.64 & 7.45 & 7.47 & 7.50 & 7.68 & 7.69 \\ & 7.45 & 7.10 & 7.56 & 7.47 & 7.52 & 7.47 \\ \hline \text { Bottled } & 5.15 & 5.09 & 5.26 & 5.20 & 5.02 & 5.23 \\ & 5.28 & 5.26 & 5.13 & 5.26 & 5.21 & 5.24 \\ \hline \end{array} $$ (a) Determine the mean, median, and mode \(\mathrm{pH}\) for each type of water. Comment on the differences between the two water types. (b) Suppose the \(\mathrm{pH}\) of 7.10 in tap water was incorrectly recorded as \(1.70 .\) How does this affect the mean? the median? What property of the median does this illustrate?

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.