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SOVA2 [1]
1 year ago
11

The number of DVDs in a random person’s home collection is counted for a sample population of 80 people. The mean of the sample

is 52 movies; the entire population is known to have a standard deviation of 12 movies. Assuming a 99% confidence level, find the margin of error.
Mathematics
1 answer:
VLD [36.1K]1 year ago
7 0

Answer:

E = 3.46\ movies

Step-by-step explanation:

The formula to find the error is:

E = z_{\frac{\alpha}{2}}\frac{\sigma}{\sqrt{n}}

Where:

\sigma is the standard deviation

n is the sample size

So

n = 80 people

\sigma = 12 movies

Then

1- \alpha = confidence level = 0.99

\alpha= 1-0.99

\alpha = 0.01\\\\\frac{\alpha}{2} = 0.005

We look for the Z value: Z_{0.005}

Z_{0.005}=2.58  Looking in the normal standard tables

Therefore:

E =2.58*\frac{12}{\sqrt{80}}\\\\E = 3.46\ movies

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To a survey of 290 newspaper readers, 181 of them read the Daily Times, 142 Read the Guardian 117 read punch and each reads at least one of the three papers if 75 read the Daily Times, and the Guardian, 60 read the Daily Times and punch, and 54 read the Guardian and punch : Draw a Venn diagram to illustrate this information How many readers read All three papers ,Exactly two of the papers The Guardian alone?

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