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**article discuss**

**es**the sample

**mean**,

**standard error**, and the

**sampling distribution**of p, the random sample with a size of 100. This article uses the concept of sample mean and standard error.

**The sample mean**is defined as the average of a set of data.

**Standard error**is how different the population mean is from the

**sample mean**.

**Expert Answer**

**Step 1**

**Consider**$p=0.40$ and $n=7$. \[E(\bar p)=p\] \[=0.40\] The

**expected value**of $\bar p$ is$0.40$.

**The expected value**$\bar p$ is the sample proportion’s mean of the

**sampling distribution.**

**Step 2**

The**standard error**of $\bar p$ is calculated as follows: \[\sigma _{\bar p}=\sqrt{\dfrac{0.40 (1-0.40)}{100}}\] \[=\sqrt {\dfrac{0.40\times 0.60}{100}}\] \[=\sqrt{0.0024}\] \[=0.0490\] The

**standard error**of $\bar p$ is $0.0490$. The

**standard error**of $\bar p $ is obtained by

**dividing the product**$p$ and $(1-p)$ by the

**sample size**$n$ and then taking the

**square root.**

**Step 3**

Check whether the **sample distribution**of $\bar p$ is

**normal.**

**Obtain**the value of $np$. \[np=100 (0.40) \] \[=40 <5\]

**Obtain**the value of $n(1-p)$ \[n(1-p)=100(1-0.40)\] \[= 60<5\] Since the values of $np$ and $n(1-p)$ are

**greater than**$5$, the sampling distribution of $\bar p$ is

**approximately normal**. The

**sample proportion’s mean**of the sampling distribution is the population proportion $p$, which is $0.40$. The

**sample proportion’s standard deviation**of the sampling distribution is $0.0490$. The

**sample distribution**of $\bar p$ has

**a mean**and

**standard deviation**is $0.40$ and $0.0490$, respectively. The

**general condition of normality**of the sampling distribution of the

**sample proportion**is fulfilled. The sampling distribution of $\bar p$ is approximately

**normal**since $np$ and $n(1−p)$ are

**greater**than $5$. By

**central limit theorem,**mean of the

**sample distribution**is equal to the proportion of the population distribution for a

**large sample.**The variance of sampling distribution is obtained from the ratio of $p(1−p)$ and the

**sample size.**

**Step 4**

According to **central limit theorem,**

**sampling distribution**of sample proportion $\bar p$ shows the probability distribution for the sample proportion. The

**sampling distribution**of sample proportion $\bar p$ shows the probability distribution for the

**sample proportion.**Sampling distribution of sample proportion is approximately normal when $n≥30$ using the

**central limit theorem**. It is important to obtain the probabilities of the

**sample proportions.**

**Numerical Results**

- The
**expected value**of $\bar p $ is $0.40$. - The
**standard error**of $\bar p$ is $0.0490$. - The
**sample distribution**of $\bar p$ is**approximately normal.**

**Example**

**A simple random sample size of 200 is selected from a population with p=0.50.What is expected value of $\bar p$? What is standard error of $\bar p$?**

**Solution**

**Step 1**

**Consider**$ p = 0.50 $ and $ n = 7$. \[E(\bar p) = p \] \[=0.50\] The

**expected value**of $\bar p$ is $0.50$.

**Step 2**

The**standard error**of $\bar p$ is calculated as follows \[\sigma _{\bar p} = \sqrt {\dfrac{0.50 (1-0.50)}{200}}\] \[ = \sqrt {\dfrac{0.50 \times 0.50 }{200}}\] \[= \sqrt {0.00125}\] \[= 0.0353\] The

**standard error**of $\bar p$ is $0.0353$.

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