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.)

(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:

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

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

Thus, according to the central limit theorem the distribution of is Normal, i.e. .

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

Compute the value of as follows:

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

Compute the value of as follows:

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

answer: l

step-by-step explanation: