Please help that’s what I came up with so far. 5. Suppose a simple random sample

Question

Please help that’s what I came up with so far. 5. Suppose a simple random sample of size n 200 is obtained from a population who size is N 25,000 and whose population proportion with a specified characteristic is p- 0.65. a) Is the sampling distribution of p approximately normal? 723 Why065 0-0337 909 b) What is the probability of obtaining x -136 or more individuals from the 200 with the characteristic? That is, what is P ( 20.68), :z003x/200. 2 p 85 1-0-8193 0bl c) What is the probability of obtaining x -118 or fewer individuals from the 200 with the characteristic?

Explanation / Answer

Solution:

Question 5

Part a

We are given

n = 200

p = 0.65, q = 1 – p = 1 – 0.65 = 0.35

Sampling distribution of p-hat is approximately normal because

n*p = 200*0.65 = 130

n*q = 200*0.35 = 70

n*p > 10 and n*q > 10

So, we can use normal approximation.

Part b

We have to find P(X136)

P(X136) = 1 – P(X<136)

For normal distribution, we have

Mean = n*p = 200*0.65 = 130

SD = sqrt(n*p*q) = sqrt(200*0.65*0.35) = 6.745369

We have to find P(X<136)

So, by subtracting continuity correction 0.5, we have to find P(X<135.5)

Z = (X – mean) / SD

Z = (135.5 – 130) / 6.745369

Z = 0.741249

P(Z<0.741249) = P(X<135.5) = 0.770729

(By using normal z-table)

P(X136) = 1 – P(X<136) = 1 - 0.770729 = 0.229271

Required probability = 0.229271

Part c

Here, we have to find P(X118)

By adding continuity correction 0.5, we have to find P(X<118.5)

Z = (118.5 - 130) / 6.745369

Z = -1.70487

P(Z<-1.70487) = P(X118) = 0.044109

Required probability = 0.044109

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.