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2 years ago

A population mean of 360, a standard deviation of 4, and a margin of error of 2.5%

Mathematics

1 answer:

Ad libitum [116K]2 years ago

3 0

Answer:

(351.04, 368.96)

Step-by-step explanation:

Given that population mean= 360

Std devitaion = 4

margin of error = 2.5%= 0.025

Since population std deviation is known we can use Z critical value

For two tailed critical value for 2.5% would be

2.24

Margin of error = ±Critical value*std deviation

= ±2.24(4)

= ±8.96

Confidence interval =(Mean - margin of error, mean + margin of error)

= $(360-8.96, 360+8,.96)\\= (351.04, 368.96)$

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2 years ago

A summer camp cookout is planned for the campers and their families. There is room for 450 people. Each adult costs $7, and each

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Answer:

$\left\{\begin{array}{l}x+y\le 450\\7x+4y\le 1,150\end{array}\right.$

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2 years ago

For Joe's illness, the doctor gave him a prescription of 28 pills. He needs to take 4 pills the first day, and then 2 pills each

kiruha [24]

Answer:

12 days.

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2 years ago

In measuring reaction time, a psychologist estimates that a standard deviation is .05 seconds. How large a sample of measurement

blondinia [14]

Answer:

Step-by-step explanation:

We are asked to find the size of sample to be 95% confident that the error in psychologist estimate of mean reaction time will not exceed 0.01 seconds.

We will use following formula to solve our given problem.

$n\geq (\frac{z_{\alpha/2}\cdot\sigma}{E})^2$ , where,

$\sigma=\text{Standard deviation}=0.05$ ,

$\alpha=\text{Significance level}=1-0.95=0.05$ ,

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$E=\text{Margin of error}$

$n=\text{Sample size}$

Substitute given values:

$n\geq (\frac{z_{0.025}\cdot\sigma}{E})^2$

$n\geq (\frac{1.96\cdot0.05}{0.01})^2$

$n\geq (\frac{0.098}{0.01})^2$

$n\geq (9.8)^2$

$n\geq 96.04$

Therefore, the sample size must be 97 in order to be 95% confident that the error in his estimate of mean reaction time will not exceed 0.01 seconds.

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2 years ago

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2 years ago

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