Question

Suppose a simple random sample of size nequals1000 is obtained from a population whose size is Nequals1 comma 000 comma 000 and whose population proportion with a specified characteristic is p equals 0.74 . Complete parts​ (a) through​ (c) below. ​(a) Describe the sampling distribution of ModifyingAbove p with caret. A. Approximately​ normal, mu Subscript ModifyingAbove p with caretequals0.74 and sigma Subscript ModifyingAbove p with caretalmost equals0.0139 B. Approximately​ normal, mu Subscript ModifyingAbove p with caretequals0.74 and sigma Subscript ModifyingAbove p with caretalmost equals0.0004 C. Approximately​ normal, mu Subscript ModifyingAbove p with caretequals0.74 and sigma Subscript ModifyingAbove p with caretalmost equals0.0002 ​(b) What is the probability of obtaining xequals770 or more individuals with the​ characteristic? ​P(xgreater than or equals770​)equals nothing ​(Round to four decimal places as​ needed.) ​(c) What is the probability of obtaining xequals720 or fewer individuals with the​ characteristic? ​P(xless than or equals720​)equals nothing ​(Round to four decimal places as​ needed.)

Answer 1

Answer:

(a) Correct option is (A).

(b) The value of P (X ≥ 770) is 0.0143.

(c) The value of P (X ≤ 720) is 0.0708.

Step-by-step explanation:

Let X = number of elements with a particular characteristic.

The variable p is defined as the population proportion of elements with the particular characteristic.

The value of p is:

p = 0.74.

A sample of size, n = 1000 is selected from a population with this characteristic.

(a)

According to the Central limit theorem, if from an unknown population large samples of sizes n > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.

The mean of this sampling distribution of sample proportion is:

 $\mu_{\hat p}=p$

The standard deviation of this sampling distribution of sample proportion is:

 $\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}$

The sample selected is of size, n = 1000 > 30.

Thus, according to the central limit theorem the distribution of $\hat p$ is Normal, i.e. $\hat p\sim N(\mu_{\hat p}=0.74,\ \sigma_{\hat p}=0.0139)$.

Thus the correct option is (A).

(b)

We need to compute the value of P (X ≥ 770).

Apply continuity correction:

P (X ≥ 770) = P (X > 770 + 0.50)

                  = P (X > 770.50)

Then $\hat p> \frac{770.5}{1000}=0.7705$

Compute the value of $P(\hat p> 0.7705)$ as follows:

$P(\hat p> 0.7705)=P(\frac{\hat p-\mu_{\hat p}}{\sigma_{\hat p}}>\frac{0.7705-0.74}{0.0139})$

                      $=P(Z>2.19)\\=1-P(Z$

Thus, the value of P (X ≥ 770) is 0.0143.

(c)

We need to compute the value of P (X ≤ 720).

Apply continuity correction:

P (X ≤ 720) = P (X < 720 - 0.50)

                  = P (X < 719.50)

Then $\hat p$

Compute the value of $P(\hat p$ as follows:

$P(\hat p$

                      $=P(Z$

Thus, the value of P (X ≤ 720) is 0.0708.