A 2007 article states that 4.8% of U.S. households are “linguistically isolated,” which means that all members of the household aged 14 years and older have difficulty speaking English (Source:...

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A 2007 article states that 4.8% of U.S. households are “linguistically isolated,” which means that all members of the household aged 14 years and older have difficulty speaking English (Source: http://www.antara.co.id/en/arc/2007/9/12/five-percent-of-us-families-dont-speak-english-report/). Assume that this percentage is true for the current population of U.S. households. Find the probability that in a random sample of 750 U.S. households, more than 45 would be classified as “linguistically isolated.”




Answered Same DayDec 25, 2021

Answer To: A 2007 article states that 4.8% of U.S. households are “linguistically isolated,” which means that...

David answered on Dec 25 2021
125 Votes
X ∼ Binomial(n = 750, p = 0.048)
Binomial can be approximated to normal with:
µ = np = 750 ∗ 0.048
= 36
σ =

np(1 − p) =

750 ∗ (0.048)(1 − 0.048) = 5.854229 ≈ 5.854
P (X > 45)
Since we are approximating a discrete binomial distribution by continuous nor-
mal distribution, values between 44.5 and 45.5 both approximate to 45. Thus,
“greater than 45 ”...
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