MATH SOLVE

4 months ago

Q:
# Playing sports related to job pay. Does participation in youth and/or high school sports lead to greater wealth later in life? This was the subject of a recent Harris POLL (March 2015). The poll found that 15% of adults who participated in sports now have an income greater than $100,000. In comparison, 9% of adults who did not participate in sports have an income greater than $100,000. Consider a random sample of 25 adults, all of whom have participated in youth and/or high school sports. a. What is the probability that fewer than 20 of these adults have an income greater than $100,000? b. What is the probability that between 10 and 20 of these adults have an income greater than $100,000? c. Repeat parts a and b, but assume that none of the 25 sam- pled adults participated in youth and/or high school sports.

Accepted Solution

A:

Answer:a) P(X<20) β
1b) P(10<X<20) = 0.000495c) P(X<20) β
1 Β P(10<X<20) = 0.000000838 β
0Step-by-step explanation:Let be the event X : ''Number of adults having an income greater than $100,000''We can modelate X as a Binomial random variableX~Bi(n,x,p)Where n is the number of the random samplex is the X value (number of success)p is the success probabilityIn our exercise :[tex]n=25\\p=0.15[/tex]The probability function of X is :[tex]P(X=x)=f(x)=nCx.p^{x}.(1-p)^{n-x}[/tex]Where [tex]nCx[/tex] is the combinatorial number : [tex]nCx=\frac{n!}{x!(n-x)!}[/tex]For a)[tex]P(X<20)=1-P(X\geq 20)=1-[f(20)+f(21)+f(22)+f(23)+f(24)+f(25)][/tex]P(X<20) = 0.9999 β
1b) [tex]P(10<X<20)=f(11)+f(12)+f(13)+f(14)+f(15)+f(16)+f(17)+f(18)+f(19)[/tex][tex]P(10<X<20)=0.000495[/tex]c) Is the same as part a) but now the probability p =0.09[tex]P(X<20) =1-P(X\geq 20)=1-[f(20)+f(21)+f(22)+f(23)+f(24)+f(25)][/tex]P(X<20) =0.9999 β
1[tex]P(10<X<20)=f(11)+f(12)+f(13)+f(14)+f(15)+f(16)+f(17)+f(18)+f(19)[/tex]P(10<X<20)=0.000000838β
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