Question

A survey of adults found that 3737​% say their favorite sport is professional football. You randomly...

A survey of adults found that

3737​%

say their favorite sport is professional football. You randomly select

120120

adults and ask them if their favorite sport is professional football.​(a) Find the probability that at most

6666

people say their favorite sport is professional football.​(b) Find the probability that more than

3333

people say their favorite sport is professional football.​(c) Find the probability that between

4343

and

5151

​people, inclusive, say their favorite sport is professional football.

​(d) Are any of the probabilities in parts​ (a)-(c) unusual? Explain.

Use a normal approximation to the binomial distribution if possible.

​(a) The probability that at most

6666

people say their favorite sport is professional football is nothing .is .

​(Round to four decimal places as​ needed.)

​(b) The probability that more than

3333

people say their favorite sport is professional football is nothing .is .

​(Round to four decimal places as​ needed.)

​(c) The probability that between

4343

and

5151

​people, inclusive, say their favorite sport is professional football is nothing .is .

​(Round to four decimal places as​ needed.)

​(d) Are any of the probabilities in parts​ (a)-(c) unusual? Explain. Select all that apply.

A.

​No, none of the probabilities is less than 0.05.

B.

​Yes, the event in part​ (b) is unusual because its probability is less than 0.05.

C.

​Yes, the event in part​ (a) is unusual because its probability is less than 0.05.

D.

​Yes, the event in part​ (c) is unusual because its probability is less than 0.05.

Homework Answers

Answer #1

Using normal approximation to the binomial distribution

mean = n*p = 0.37*120 = 44.4

standard deviation = sqrt(n*p*(1-p))

=sqrt(120*0.37*0.63)

=5.29

(A) using normalcdf

setting

lower = -99

upper = 66+0.5..........(0.5 due to continuity correction)

mean = 44.4

standard deviation= 5.29

we get

P(at most 66) = normalcdf(-99,66.5,44.4,5.29)

= 1.0000

(B) using normalcdf

setting

lower = 33+0.5..........(0.5 due to continuity correction)

upper = 99

mean = 44.4

standard deviation= 5.29

we get

P(more than 33) = normalcdf(33.5,99,44.4,5.29)

= 0.9803

(C)

using normalcdf

setting

lower = 43-0.5..........(0.5 due to continuity correction)

upper = 51+0.5.........(0.5 due to continuity correction)

mean = 44.4

standard deviation= 5.29

we get

P(between 43 and 51) = normalcdf(42.5,51.5,44.4,5.29)

= 0.5506

(D) we can see that none of the probabilities is less than 0.05

so, there is no unusual event

option A

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