susana benitez | 1/24/18 3:03 PM Test: Ch 4 Test Time Limit: 02:00:00 Submit Tes
ID: 3352902 • Letter: S
Question
susana benitez | 1/24/18 3:03 PM Test: Ch 4 Test Time Limit: 02:00:00 Submit Test This Question: 1 pt | 14of14(0 complete) This Test: 14 pts possible ar Use technology (such as a binomial calculator) to find the probabilities shown below. Then determine if the events are unusual. reasoning. Explain your Sixty-seven percent of pet owners say they consider their pet to be their best friend. You randomly select 10 pet owners and ask them if they consider their pet to be their best friend. Find the probability that the number who say their pet is their best friend is (a) exactly eight, (b) at least seven, and (c) at most three. (a) P(exacty eight) (Round to four decimal places as needed.) ok e (b) P(at least seven)-Round to four decimal places as needed,) (c) P(at most three- (Round to four decimal places as needed.) Which of the events are unusual? Select all that apply. A. The event in part (a) is unusual because it is highly unlikely to occur. B. The event in part (b) is unusual because it is highly unlikely to occur. C. D. The event in part (c) is unusual because it is highly unlikely to occur None of the events are unusual because none of them are highly unlikely to occur.Explanation / Answer
p=67% =0.67
q=1-p =1-0.67
=0.33
n=10
a)
Here r=8
P=nCr p^r q^(n-r)
=10C8 (0.67)^8 (0.33)^(10-8)
=10C8 (0.67)^8 (0.33)^2
=0.1990
b)
P(at least seven) =P(r>=7)
=P(r=7) +P(r=8) + P(r=9) +P(r=10)
=10C7 (0.67)^7 (0.33)^(3) +10C8 (0.67)^8 (0.33)^(2) +10C9 (0.67)^9 (0.33)^(1) +10C10 (0.67)^10 (0.33)^(0)
=0.261364 +0.19899 +0.08978 +0.0182
=0.5684
c)
P(at most 3) = P(r<=3)
=BINOM.DIST(3,10,0.67,TRUE) =0.0185
Since probability for part C is less than 0.05,hence this is unusual event
so its choice C
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