Explanation Given information: In 2010 , there are 71 % of all the households have internet access in the United States and there are total of 1 , 200 households in the town. Formula used: Normal approximation of a binomial random variable X with n independent trials with p as the probability of success is given by, P ( X ≥ b ) ≈ P ( Y ≥ a − 0.5 ) Where Y is a normal random variable with similar mean and standard deviation as X and is given by, μ = n p σ = n p q Where, q is the probability of the failure. The standard form of probability P ( X ≥ a ) when X has a normal distribution with mean μ and standard deviation σ is P ( Z ≥ a − μ σ ) where, Z is the standard normal variable. Calculation: Consider the binomial random variable, X which represents the number of households have internet access from the total of 1 , 200 households. The probability for the binomial random variable X is P ( X ≥ 840 ) . Normal distribution of binomial random variable X is given by, P ( X ≥ 840 ) ≈ P ( Y ≥ 840 − 0.5 ) = P ( Y ≥ 839.5 ) Here, n = 1 , 200 and the probability of success that a household have an internet access is p = 0.71 . As p = 0.71 , the value of q is calculated as, q = 1 − p = 1 − 0.71 = 0.29 Now, calculate the value of mean μ by the use of the above formula as, μ = n p = 1 , 200 ⋅ 0.71 = 852 Then, calculate the value of standard deviation σ as, σ = n p q = 1 , 200 ⋅ 0

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Chapter 8.5, Problem 57E

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To calculate: The probability that the number of households have internet access in 2010 is at least 840 from the total of 1,200 households in the town and in 2010, there are 71% of all the households have internet access in the United States.

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Chapter 8.5, Problem 56E

Chapter 8.5, Problem 58E

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The probability that the number of households have internet access in 2010 is at least 840 from the total of 1 , 200 households in the town and in 2010 , there are 71 % of all the households have internet access in the United States.

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To calculate: The probability that the number of households have internet access in 2010 is at least 840 from the total of 1,200 households in the town and in 2010, there are 71% of all the households have internet access in the United States.

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