Give a typed answer Suppose that 53% of all registered voters prefer presidentia

Question

Give a typed answer

Suppose that 53% of all registered voters prefer presidential candidate Smith to presidential candidate Jones. (You can substitute the names of the most recent presidential candidates.)

a. In a random sample of 100 voters, what is the probability that the sample will indicate that Smith will win the election (that is, there will be more votes in the sample for Smith)?

b. In a random sample of 100 voters, what is the probability that the sample will indicate that Jones will win the election?

c. In a random sample of 100 voters, what is the probability that the sample will indicate a dead heat (fifty-fifty)?

d. In a random sample of 100 voters, what is the probability that between 40 and 60 (inclusive) voters will prefer Smith?

Explanation / Answer

Solution:

Part a

Here, we have to use normal approximation to binomial distribution. We are given n = 100 and p = 0.53, q = 1 – p = 1 – 0.53 = 0.47

So we have,

Mean = n*p = 100*0.53 = 53 and

SD = sqrt(n*p*q) = sqrt(100*0.53*0.47) = 4.990992 or 5 approximately

Here, we have to find the probability that the sample will indicate that Smith will win.

This means we have to find P(X>50)

By adding continuity correction 0.5, we have to find P(X>50+0.5) = P(X>50.5)

P(X>50.5) = 1 – P(X<50.5)

Z = (X – mean) / SD

Z = (50.5 – 53) / 5 = -0.5

P(X<50.5) = P(Z<-0.5) = 0.308538

P(X>50.5) = 1 – P(X<50.5) = 1 - 0.308538 = 0.691462

Required probability = 0.691462

Part b

Here, we have to find the probability that the sample will indicate that Jones will win.

This means we have to find P(X>50)

For Jones, winning probability = 1 – 0.53 = 0.47

So, mean = n*p = 100*0.47 = 47

SD = sqrt(npq) = sqrt(100*0.47*0.53) = 5 approximately

By adding continuity correction 0.5, we have to find P(X>50+0.5) = P(X>50.5)

P(X>50.5) = 1 – P(X<50.5)

Z = (50.5 – 47)/5 = 0.7

P(X<50.5) = P(Z<0.7) = 0.758036

P(X>50.5) = 1 – 0.758036 = 0.241964

Required probability = 0.241964

Part c

Here, we have to find P(X=50)

By adding and subtracting continuity correction, we have to find

P(49.5<X<50.5) = P(X<50.5) – P(X<49.5)

P(X<50.5) = P(Z<-0.5) = 0.308538

Now, we have to find P(X<49.5)

Z = (49.5 – 53)/5 = -0.7

P(X<49.5) = P(X<-0.7) = 0.241964

P(49.5<X<50.5) = P(X<50.5) – P(X<49.5)

P(49.5<X<50.5) = 0.308538 - 0.241964 = 0.066574

Required probability = 0.066574

Part d

Here, we have to find P(40<X<60)

n = 100 and p = 0.53, q = 1 – p = 1 – 0.53 = 0.47

Mean = n*p = 100*0.53 = 53 and

SD = sqrt(n*p*q) = sqrt(100*0.53*0.47) = 4.990992 or 5 approximately

P(40<X<60) = P(X<60) – P(X<40)

First we have to find P(X<60)

Z = (60 – 53)/5 = 1.4

P(X<60) = P(Z<1.4) = 0.919243

Now, we have to find P(X<40)

Z = (40 – 53)/5 = -2.6

P(X<40) = P(Z<-2.6) = 0.004661

P(40<X<60) = P(X<60) – P(X<40)

P(40<X<60) = 0.919243 - 0.004661 = 0.914582

Required probability = 0.914582

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