Question

A binomial distribution has

pequals=0.550.55

and

nequals=4040.

A binomial distribution has p=0.550.55 and n=4040.

a. What are the mean and standard deviation for this​ distribution?

b. What is the probability of exactly 24 ​successes?

c. What is the probability of fewer than 27 ​successes?

d. What is the probability of more than19 ​successes?

Answer 1

Since n is very large, we use the binomial approximation to the normal for the normal.

When using the Normal approximation, we use the continuity correction factor, the table for which is given below.

Here n = 40, p = 0.55, q = 1 – p = 0.45

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(a) Mean = n * p = 40 * 0.55 = 22

Standard Deviation = Sqrt(n * p * q) = Sqrt(40 * 0.55 * 0.45) = 3.146

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(b) X = 24, Using the continuity correction factor, we need to find P(23.4 < X < 24.5) = P(X < 24.5) - P(X < 23.5)

For P(X < 24.5), Z = (24.5 - 22) / 3.146 = 0.79

For P(X < 23.5), Z = (23.5 - 22) / 3.146 = 0.48

The required probability = P(Z = 0.79) - P(Z = 0.48) = 0.7852 - 0.6844 = 0.1008

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(c) P(X < 27), Using the continuity correction factor, we need to find P(X < 26.5)

For P(X < 26.5), Z = (26.5 - 22) / 3.146 = 1.53

The required probability = 0.9236

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(d) P(X > 19), Using the continuity correction factor, we need to find P(X > 19.5) = 1 - P(X < 19.5)

For P(X < 19.5), Z = (19.5 - 22) / 3.146 = -0.79

The probability for P(X < 19.5) = 0.2148

The required probability = 1 - 0.2148 = 0.7852

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The continuity correction factor table

If P(x = n), then use P(n – 0.5 < X < n + 0.5)

If P(X > n) , then use P(X > n + 0.5)

If P(X < n) , then use P(X < n - 0.5)

If P(X n) , then use P(X > n - 0.5)

If P(X n) , then use P(X < n + 0.5)

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