Question

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.

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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