Step E: Probability and the Binomial Distribution As you know from a good readin

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Step E: Probability and the Binomial Distribution As you know from a good reading of our text, we can estimate probabilities and cumulative probabilities by using the "normal approximation to the binomial distribution." That is, we can use the Unit Normal Table to tell us the probabilities of certain outcomes of a process having two outcomes (like success or failure, or heads vs. tails). Consider the distribution of outcomes of 76 coin flips. Does this meet the requirement for using the normal approximation to the binomial? Explain briefly how the criteria are (or are not) met. (Calculate pn and qn) What is the mean of this distribution, the average number of "heads" in 76 coin flips? Type your answer in cell What is the standard deviation? Type your answer in cell B67. What is the probability of a getting 28 or fewer heads in 76 coin flips? This takes a few steps. First, remember that the real limits of 28 in a continuous distribution, which is what the normal distribution is, are 27.5, 28.5. So use the ceiling of this range (X=28.5) in your z-score calculation. Calculate the z-score and round to 2 decimal places. Look up the z-score you calculated in Table B. 1. What is this probability?

Explanation / Answer

n = 76 , p = 1/2 , q = 1/2 , probability of getting head = probability of getting tail

1)

a) np > 5 , nq> 5 , requirement for using normal approximation to binomial .

clearly np = nq = 38 > 5

b) mean = np = 76 * 1/2 = 38

2) variance = np(1-p) = 78*1/2*1/2 = 19

sd= sqrt(variance) = sqrt(19) =4.3588989435

3) P(X <=28)

X is N(38,19)

using continuity Correction Factor

P(X <=28) = P(X<28.5)

b) Z =(X- 38)/4.3588989435

(28.5-38)/4.3588989435

=-2.1794494

c) P(Z< -2.1794494) = 0.0146

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