1. Let X - the number of textbooks purchased by fulltime UofA s probability dist
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Question
1. Let X - the number of textbooks purchased by fulltime UofA s probability distribution of X. tudents this semester. The following is a valid P(X) 01 .05 06 10 70 04 P(x-2) How many textbooks, on average, did University of Akron students purchased this semester? 2. Indicate if the following is a binomial experiment. Put 'yes' if binomial and 'no' if not. a.) For 10 human births, it was recorded whether the births took more than or (9 months or less) for the baby to arrive The probability that a human gestation period will exceed 9 months is 30%. b.) For 10 human births, the actual amount of time for the baby to arrive was recorded. c.) A company tests a golf club by using a robot to hit a golf ball 12 times with a head velocity of 85 miles per hour. The total yards each ball traveled is recorded. d) A 6-sided die is rolled 18 times and the % of times each number is rolled is calculated. e.) A 6-sided die is rolled 18 times and the number of times an odd is rolled is recorded. 3. Suppose a cereal manufacturer put pictures of famous athletes on cards in boxes of cereal, in the hope of increasing sales. The manufacturer announces that 40% of the boxes contain a card with Tiger Woods's picture. The other 60% contain pictures of other athletes. Mom has a coupon that allows her to buy 5 boxes at a discounted amount, so she decides to buy 5 boxes. What is the probability that after we opened them: aj exactly 2 of them will have a picture of Tiger Woods? P(x 2) 0433 b) none of them have a picture of Tiger woods? PX :0) Hint: the answer is NOT zero. D c.) all the boxes have a picture of Tiger Woods? P(X- 5) d.) 1 to 4 boxes have a picture of Tiger Woods? P(1 sXs4) e.) What is the average number of boxes that will have Tiger's picture? Hint: calculate np f.) Standard deviation of the number of boxes with Tiger's picture? Hint: calculate np(l-p)Explanation / Answer
Solution(1)
from the given table we can found that probability of P(X=2) = 1-P(X=0)-P(X=1)-P(X=3)-P(X=4)-P(X=5)-P(X=6) = 1- 0.01-0.05-0.06-0.7-0.1-0.04 = 1-0.96 = 0.04
P(X<2) = P(X=0)-P(X=1) = 0.01+0.05 = 0.06
P(X <=2) = P(X=0)+P(X=1)+P(X=2) = 0.01+0.05+0.04 = 0.1
P(X>5) = P(X=6) = 0.04
P(1<X<=5) = P(X=2)+P(X=3)+P(X=4)+P(X=5) = 0.04+0.06+0.1+0.7 = 0.9
Average books = (0*0.01) +(1*0.05) +(2*0.04) +(3*0.06) +(4*0.1)+(5*0.7)+(6*0.04) = 0+0.05+0.08+0.18+0.4+3.5+0.24 = 4.45
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