Answer:
For the sampling distribution of the sample proportion for a sample of size 50, the mean is 0.061 and the standard deviation is 0.034.
Explanation:
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean
and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean
and standard deviation
![s = \sqrt{(p(1-p))/(n)}](https://img.qammunity.org/2021/formulas/mathematics/college/4g01jif87kw0yiycg79zy61z1uo268l9th.png)
In this question:
![p = 0.061, n = 50](https://img.qammunity.org/2021/formulas/mathematics/college/195br6tgagkgbkbbcp2xgrlk9b480s42fu.png)
So
Mean:
![\mu = p = 0.061](https://img.qammunity.org/2021/formulas/mathematics/college/8t0td21ggczi44dt6bx1qsxp8a2e7jlvi2.png)
Standard deviation:
![s = \sqrt{(0.061*0.939)/(50)} = 0.034](https://img.qammunity.org/2021/formulas/mathematics/college/4u5k2rsqv70qooj4558qsipqd2f4yx3om9.png)
For the sampling distribution of the sample proportion for a sample of size 50, the mean is 0.061 and the standard deviation is 0.034.