Probability and the Binomial Distribution As you know from a good reading of our

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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.

1. Does this meet the requirement for using the normal approximation to the binomial? Explain briefly how the criteria are (or are not) met.

2. What is the mean of this distribution, the average number of "heads" in 76 coin flips?

3. What is the standard deviation?

4. 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.

5. Look up the z-score you calculated in Table B.1. What is this probability? Type your answer (as shown in table).

Test Score z-Score 45 -1.317 48 -1.135 51 -0.953 51 -0.953 52 -0.892 52 -0.892 53 -0.831 53 -0.831 53 -0.831 54 -0.770 54 -0.770 55 -0.709 57 -0.588 58 -0.527 59 -0.466 59 -0.466 68 0.081 69 0.142 69 0.142 70 0.203 75 0.507 77 0.628 78 0.689 85 1.115 85 1.115 88 1.297 93 1.601 94 1.662 96 1.784 99 1.966 Mean 66.67 Median 59 Mode 53 SD - Sample 16.44705892 SD - Population 16.17061807

Explanation / Answer

1) here p=0.5 and n=76

hence np=38=n(1-p)>10 ; therefore meets the requirement

2)mean =np=38

3) std deviation =(np(1-p))1/2=4.3589

4) here P(X<=28) =P(Z<(28.5-38)/4.3589)=P(Z<-2.1794) (Z score =-2.1794)

5) =0.0146

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