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The gestation time of humans has an approximate Normal distribution with a mean of 250 days and a standard deviation of 6.0 days. A simple random sample of n newborns is to be taken.
What is the minimum sample sized needed so that the sampling distribution of bar(x) has a standard deviation of 0.5 day?
(A) 15
(B) 144
(C) 12
(D) 250

The gestation time of humans has an approximate Normal distribution with a mean of $250$ days and a standard deviation of $6.0$ days. A simple random sample of $n$ newborns is to be taken. $\newline$ What is the minimum sample size $n$ needed so that the sampling distribution of $\bar{x}$ has a standard deviation of $0.5$ day? $\newline$ (A) $15$ $\newline$ (B) $144$ $\newline$ (C) $12$ $\newline$ (D) $250$

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Interpret measures of center and variability

Full solution

Q. The gestation time of humans has an approximate Normal distribution with a mean of

250

days and a standard deviation of

6.0

days. A simple random sample of

n

newborns is to be taken.

\newline

What is the minimum sample size

n

needed so that the sampling distribution of

\bar{x}

has a standard deviation of

0.5

day?

\newline

(A)

15

\newline

(B)

144

\newline

(C)

12

\newline

(D)

250

Understand Formula: Understand the formula for the standard deviation of the sampling distribution of the mean, which is the standard error of the mean (SEM). $\newline$ $\text{SEM} = \frac{\sigma}{\sqrt{n}}$ , where $\sigma$ is the population standard deviation and $n$ is the sample size.
Plug in Values: Plug in the given values into the formula to find $n$ . We know $\sigma = 6.0$ days and we want the SEM to be $0.5$ days. $0.5 = \frac{6.0}{\sqrt{n}}$
Solve for $n$ : Solve for $n$ . $\newline$ Square both sides of the equation to eliminate the square root: $\newline$ $(0.5)^2 = (6.0 / \sqrt{n})^2$ $\newline$ $0.25 = 36 / n$
Rearrange Equation: Rearrange the equation to solve for $n$ . $n = \frac{36}{0.25}$ $n = 144$

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