Mathematics, 14.04.2020 16:30 brianamarialove15

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.)

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Answer from: sineguguhopengw3912

(a) Correct option is (A).

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

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:

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

$=P(Z2.19)\\=1-P(Z$

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.

Answer from: Quest

step-by-step explanation:

12: 00 p.m. to 7: 00 p.m. is a total of 7 hours.

4: 00 p.m. to 7: 00 p.m. is a total of 3 hours.

this means that from 4: 00 p.m to 7: 00 p.m. this represents 3/7 of what was lost from 12: 00 p.m. to 7: 00 p.m.

after knowing this, after doing 11,600 - 6,800 = 4,800

4,800 represents 3/7 of what was lost for 7 hours.

we then divide 4,800 by 3 which gets 1,600.

because we know that 1,600 represents 1/7 of how much water was lost, we just have to multiply by 7, which gets 11,200. after adding 11,200 to how much water was left at 7: 00 p.m. which was 6,800. 11,200 + 6,800 = 18,000

this means that the pool originally had 18,000 gallons in the pool.

you can also check your work by doing 18,000 - 4(1600), which becomes 1800 - 6400 which gets you 11,600.

Answer from: Quest

i believe that the slope of the line is -5

step-by-step explanation: