loneworka 4409157138questionid-148Mushed-fase8ecld Math 1530-W2, Fall 2017 Homew

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loneworka 4409157138questionid-148Mushed-fase8ecld Math 1530-W2, Fall 2017 Homework: MSL 2 Score: 0 of 1 pt 14of25(13conpite) Score: 48%, 12 of 2S 5.6.179 EQuestion Hep The prebability is 0.4 that a taffic fatality inwolves an intoxicated or alcoho Y, which invove an intoxicated or alcchol-impaired driver or nonoccupant is a. exactly three, at least twee, at most three b. between two and four, inclunive c. Find and interpret the mean of the random variablo Y ngpatod aier ar nonocupant h nine tatie lll tht the nunt a. The probabiliny that exactly thee traffic falities involve Round to four decimal places as needed) an intoxicated or akcehal inpaired driver or nonoccupant is 2508 The probatley hat at least tree raffc fatalities involve an itmicated er alcohdinpaired diver Round to four decimal placas as needed) atmoupat s Enter your arower in the answer tox and then cick Check Ansines remaning D Type here to search DOLL 5 7 8 WE

Explanation / Answer

"Success" = fatality, p=0.4, n=9
Binomial Probability Formula:
p(Y=y) = [n!/(y!(n-y)!)] • p^y • (1-p)^n-y

A). Exactly 3: p(Y=3) = 0.2508

BPF: p(Y=3) = 9!/[3!(9-3)!] • (0.4^3) • (0.6^6)

At least 3: = 0.7682
Find p(Y=0,1,2) and subtract from 1

At most 3: p(Y=0,1,2,3) = 0,4826

B). p(Y=2-4) = 0.6629
Find p(Y=2,3,4)

C). µ = 3.6
µ of BRV = (n # of trials)(p) = (9)(0.4)
On average, of 9 traffic fatalities, this many will involve an intoxicated or alcohol-impaired driver or nonoccupant.

D). =1.47
Standard Deviation of Binomial Random Variable:
= {n • p • (1-p)}
= {9 • 0.4 • 0.6}

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