Question

According to the U.S. Department of Commerce, 30% of Arkansas households had a computer in 2002. If 285 Arkansas households are selected at random, find the following probabilities: (Please show all necessary justifications.) a. The exact probability that 70 or fewer have a computer b. The probability that 70 or fewer have a computer using the normal approximation. c. The probability that 70 or fewer have a computer using the normal approximation with the continuity correction.

Answer #1

a)

X ~ Bin ( n , p)

Where n = 285 , p = 0.30

Binomial probability distribution is

P(X) = nCx * px * ( 1 - p)n-x

Using EXCEL,

P(X <= 70) = BINOM.DIST ( number_s , trials, probability_s , cumulative )

= BINOM.DIST ( 70 , 285 , 0.30 , TRUE)

= **0.0247**

b)

Mean = n p = 285 * 0.30 = 85.5

Standard deviation = sqrt [ n p ( 1 - p) ] = sqrt [ 285 * 0.3 * ( 1 - 0.3) ] = 7.7363

Using normal approximation ,

P(X < x) = P(Z < ( x - mean) / SD )

So,

P(X <= 70) = P(Z < ( 70 - 85.5) / 7.7363)

=P (Z < -2.00)

= **0.0228**

c)

Using continuity correction

P(X <= 70) = P(X < 70.5)

= P(Z < ( 70.5 - 85.5) / 7.7363)

= P(Z < -1.94)

= **0.0262**

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