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Question

A binomial probability distribution has p= 0.20 and n = 100.

a)What is the mean and standard deviation?

b)Can you approximate these Binomial probabilities by the normal probabilities? Explain.

c)What is the probability of more than 24 successes?

d)What is the probability of 18 to 22 successes?

Answers

a.) The formula to calculate the mean of Binomial Probability Distribution is

μ = n.p

= 100 X .2

= 20

The formula to calculate the Standard Deviation of a Binomial Probability Distribution is

σ = √np.(1-p)

= √100 X .2 X .8

= 4

b.) Since the binomial distribution approaches the normal distribution as the number of trials increases, we can use the normal distribution to approximate binomial probabilities. These values will all be approximations; we could use the binomial probability to obtain the actual answers, but in some cases these probabilities are difficult to calculate by hand.

We can use the standard normal distribution with the following conversion formula: Z = (X - np)/ sqrt(npq)

c.) Probability of more than 24 successes can be phrased as

P(X>24)

Z value of X = (X - μ)/ σ

= (24.5 – 20)/4

= 1.125 ~ 1.13

= 0.3708

.3708 is the probability between the mean (20) and 24.

P(X>24) = 0.5 - .3708 = 0.1292

d.) Probability of 18 to 22 successes can be formulated as

P(18<=X<=22)

Z value= (X - μ)/ σ

= (22 – 20)/4

= .5

P(22) = 0.1915

P(18)= (18 – 20)/4

= -.5

Z Value= 1 – 0.1915

= 0.8085

Therefore P(18<=X<=22) = P(22)+ P(18)

= .1915 + 8085

= 1

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