А,

The mean percentange of a population of people eating out at least once a week is 57℅

Sidana [21]

Answer:

<u>The correct answer is B. between 56.45% and 57.55% </u>

Complete statement and question:

The mean percentage of a population of people eating out at least once a week is 57% with a standard deviation of 3.50%. Assume that a sample size of 40 people was surveyed from the population a significant number of times. In which interval will 68% of the sample means occur?

between 55.89% and 58.11%

between 56.45% and 57.55%

between 56.54% and 57.46%

between 56.07% and 57.93%

Source: brainly.com/question/1068489

Step-by-step explanation:

1. Let's review the information given to us to answer the question correctly:

Mean percentage of a population of people eating out at least once a week = 57%

Standard deviation = 3.5%

Sample size = 40

Confidence level = 68%

2. In which interval will 68% of the sample means occur?

For answering this question, we should find out the standard deviation of the sample, using this formula:

Standard deviation of the sample = Standard deviation of the population/√Sample size

Standard deviation of the sample = 3.5/√40

Standard deviation of the sample = 3.5/6.32

Standard deviation of the sample = 0.55

Let's recall that a confidence level of 68% means that 68% of the sample data would have a value between the mean - 1 time the standard deviation of the sample and the mean + 1 time the standard deviation of the sample. Thus:

57 - 1 * 0.55 = 57 - 0.55 = 56.45

57 + 1 * 0.55 = 57 + 0.55 = 57.55

<u>The correct answer is B. between 56.45% and 57.55% </u>

7 0

2 years ago

Long division with 489

RUDIKE [14]

Answer:

9.4

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

How can I measure a 173 and a 68 degrees Angle in a circle.

kifflom [539]

Using a Protractor..
It's a little plastic half circle thingy with angle measurements on it.
see image below

6 0

3 years ago

The 1996 gss asked, "if the husband in a family wants children, but the wife decides that she does not want any children, is it

zzz [600]

The confidence interval would be
$0.80\pm0.03$ .

We use the formula
$p\pm z*\sigma_p$ , where

$\sigma_p=\sqrt{\frac{p(1-p)}{N}}$ , with p being the sample proportion and N being the sample size.

First we find the z-score associated with this level of confidence:
Convert 95% to a decimal: 95/100 = 0.95
Subtract from 1: 1-0.95 = 0.05
Divide by 2: 0.05/2 = 0.025
Subtract from 1: 1-0.025 = 0.975

Using a z-table (http://www.z-table.com) we see that this value is associated with a z-score of 1.96.

Since 578/720 said yes, this gives us p=0.80:
$\sigma_p=\sqrt{\frac{0.80(1-0.80)}{720}}=\sqrt{\frac{0.8(0.2)}{720}=0.015$

This gives us
$0.80\pm1.96(0.015)=0.80\pm 0.03$

5 0

3 years ago

How to do 3/2b+1/5b=17 ?

guajiro [1.7K]

Answer:

b = 10

Explanation:

Lets start by combining like terms (b):

3/2b + 1/5b

to combine these fractions you must get them to both have the same denominator:

3/2 · 5/5 = 15/10 multiply top and bottom by other fractions denominator

1/5 · 2/2 = 2/10 multiply top and bottom by other fractions denominator

15/10 + 2/10 = 17/10

17/10b = 17

Now we want to get the b to be by itself, so lets first multiply both sides by 10:

17b = 170

Now lets divide both sides by 17:

b = 10

Hope this helped!

7 0

3 years ago

Read 2 more answers