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

(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.

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{(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> (770.5)/(1000)=0.7705`

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

`P(\hat p> 0.7705)=P((\hat p-\mu_(\hat p))/(\sigma_(\hat p))>(0.7705-0.74)/(0.0139))`

`=P(Z>2.19)\\=1-P(Z<2.19)\\=1-0.98574\\=0.01426\\\approx0.0143`

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<(719.5)/(1000)=0.7195`

Compute the value of
`P(\hat p<0.7195)` as follows:

`P(\hat p<0.7195)=P((\hat p-\mu_(\hat p))/(\sigma_(\hat p))<(0.7195-0.74)/(0.0139))`

`=P(Z<-1.47)\\=1-P(Z<1.47)\\=1-0.92922\\=0.07078\\\approx0.0708`

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