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76.9%
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Approximately
20% of newborns are born more than 1 week before their due date. A random sample of 20 newborns is selected.
The standard deviation of the sampling distribution for the proportion of your sample that is born more than 7 days before their due date is
0.20 .
3.2 .
0.089 .
0.008 .

$76.9 \%$ $\newline$ Resources $\newline$ Check $\newline$ Approximately $20 \%$ of newborns are born more than $1$ week before their due date. A random sample of $20$ newborns is selected. $\newline$ The standard deviation of the sampling distribution for the proportion of your sample that is born more than $7$ days before their due date is $\newline$ $0$ . $20$ . $\newline$ $3$ . $2$ . $\newline$ $0$ . $089$ . $\newline$ $0$ . $008$ .

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Use normal distributions to approximate binomial distributions

Full solution

Q.

76.9 \%

\newline

Resources

\newline

Check

\newline

Approximately

20 \%

of newborns are born more than

1

week before their due date. A random sample of

20

newborns is selected.

\newline

The standard deviation of the sampling distribution for the proportion of your sample that is born more than

7

days before their due date is

\newline

0

.

20

.

\newline

3

.

2

.

\newline

0

.

089

.

\newline

0

.

008

.

Use Formula: To find the standard deviation of the sampling distribution for the proportion, we use the formula for the standard deviation of a proportion, which is $\sqrt{p(1-p)/n}$ , where $p$ is the proportion and $n$ is the sample size.
Plug Values: Given that $p = 20\%$ or $0.20$ and $n = 20$ , we plug these values into the formula: $\sqrt{0.20(1-0.20)/20}$ .
Calculate Inside: Calculate the inside of the square root: $0.20 \times 0.80 / 20 = 0.016 / 20$ .
Divide by $20$ : Now divide $0.016$ by $20$ : $0.016 / 20 = 0.0008$ .
Take Square Root: Finally, take the square root of $0.0008$ : $\sqrt{0.0008} = 0.0283$ , but we round to three decimal places, so the standard deviation is $0.028$ .

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