Please show how you get Part 2, I\'m not understanding mostly c and d. Part 1: W

Question

Please show how you get Part 2, I'm not understanding mostly c and d.

Part 1: Wishing to test the claim that a certain coin is “fair”: You toss the coin 250 times, and get 140 heads. (ALREADY FOUND PART 1, NEED HELP WITH PART 2) With these given values (I already found in part one):

a.) The mean and standard deviation of the sample heads count X, for the 250 tosses. Answer: 125, Standard Deviation is approx. 7.9057%

b.) The difference d between the heads count X from your test, and the expected heads count. Answer: The difference is 15

c.) The percent probability that X would fall at least d away from expected, in the positive direction. Answer: is approx. 0.834%

d.) The percent probability that X would fall at least d away from expected, in either direction. Answer: is approx. 0.2015%

Now, what I am having trouble with is with part 2:

Part 2: Problem Statement: Refer again to the completed coin-toss experiment data in part 1 above. Using the Central Limit Theorem and the appropriate Normal Distribution Model for a fair coin, find the following values:

a.) The mean and standard deviation of the sample proportion hat{p} of heads, for the 250 tosses.

b.) The difference d between the heads proportion hat{p} from your test, and the expected heads proportion.

c.) The percent probability that hat{p} would fall at least d away from expected, in the positive direction.

d.) The percent probability that hat{p} would fall at least d away from expected, in either direction.

Explanation / Answer

the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
a.
proportion ( p ) = 0.5
standard Deviation ( sd )= sqrt(PQ/n) = sqrt(0.5*0.5/250)
=0.032
b.
the diffrence between p and expected p is = 140/250 - 0.5 = 0.06
c.
P(X < 0.56) = (0.56-0.5)/0.03162
= 0.06/0.03162= 1.89753
= P ( Z <1.89753) From Standard Normal Table
= 0.97112
P(X > = 0.56) = (1 - P(X < 0.56)
= 1 - 0.97112 = 0.02888
d.
BETWEEN THEM
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 0.44) = (0.44-0.5)/0.03162
= -0.06/0.03162 = -1.89753
= P ( Z <-1.89753) From Standard Normal Table
= 0.02888
P(X < 0.56) = (0.56-0.5)/0.03162
= 0.06/0.03162 = 1.89753
= P ( Z <1.89753) From Standard Normal Table
= 0.97112
P(0.44 < X < 0.56) = 0.97112-0.02888 = 0.94224

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