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DanielleElmas [232]
3 years ago
5

could someone help with the steps! In a survey conducted in an office, 60% of the sample respondents agreed that employees shoul

d have flexible working hours. If the margin of error is ±2%, the expected population proportion that wants flexible working hours is between ? % and ? %
Mathematics
2 answers:
nignag [31]3 years ago
7 0

Answer: The expected population proportion that wants flexible working hours is between <u> 58%</u> and <u>62%.</u>

Step-by-step explanation:

The margin of error (E) is used to represent the level of accuracy of any random sample.

Confidence interval for population percentage (p):

 (p-E, p+E ) , where E is the margin of error ( in percentage.)

Given : The percentage of the sample respondents agreed that employees should have flexible working hours : p = 60%

The margin of error : E=  ± 2%

Then , the expected population proportion that wants flexible working hours is between : 60%-2% and 60% +2%

= 58% and 62%.

Hence, the expected population proportion that wants flexible working hours is between <u> 58%</u> and <u>62%</u>.

Ira Lisetskai [31]3 years ago
5 0
58%-62%
just add 2 to 60 
and subtract 2 from 60 to find the range of percents
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Step-by-step explanation:

Let's define x as the number of cookies that you eat, and y as the number of calories you consume in those cookies.

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This means that x can be equal to 2, or larger than 2.

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3 years ago
An HP laser printer is advertised to print text documents at a speed of 18 ppm (pages per minute). The manufacturer tells you th
anastassius [24]

Answer:

0.288

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

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For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

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This is 1 subtracted by the pvalue of Z when X = 18.06. So

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Z = 0.56

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The answer is 0.288

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