1. Google it: According to a recent report, 68% of Internet searches in a particular month used the Google search engine. Assume that a sample of 22 searches is studied. Round the answers to four...

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1. Google it: According to a recent report, 68% of Internet searches in a particular month used the Google search engine. Assume that a sample of 22 searches is studied. Round the answers to four decimal places. (a) What is the probability that exactly 19 of them used Google? (b) What is the probability that 14 or fewer used Google? (c) What is the probability that more than 19 of them used Google? (d) Would it be unusual if fewer than 12 used Google? The probability is ____ 2. College bound: A national college researcher reported that 64% of students who graduated from high school in 2012 enrolled in college. Thirty two high school graduates are sampled. Round the answers to four decimal places. (a) What is the probability that exactly 20 of them enroll in college? (b) What is the probability that more than 14 enroll in college? (c) What is the probability that fewer than 14 enroll in college? (d) Would it be unusual if more than 23 of them enroll in college? The probability is ___ 3. College bound: A national college researcher reported that 64% of students who graduated from high school in 2012 enrolled in college. Thirty high school graduates are sampled. (a) What is the mean number who enroll in college in a sample of 30 high school graduates? Round the answer to two decimal places. (b) What is the standard deviation of the number who enroll in college in a sample of 30 high school graduates? Round the answer to four decimal places. 4. Use the Cumulative Normal Distribution Table to find the z-score for which the area to its left is 0.69 5. Use the Cumulative Normal Distribution Table to find the z-score for which the area to its right is 0.73. 6. Use the Cumulative Normal Distribution Table to find the z-scores that bound the middle 82% of the area under the standard normal curve. Enter the answers in ascending order. Round the answers to two decimal places. 7. Big chickens: According to a poultry industry news website, the weights of broilers (commercially raised chickens) are approximately normally distributed with mean 1376 grams and standard deviation 183 grams. Use the Cumulative Normal Distribution Table to answer the following. (a) What proportion of broilers weigh between 1105 and 1300 grams? (b) What is the probability that a randomly selected broiler weighs more than 1590 grams? (c) Is it unusual for a broiler to weigh more than 1750 grams? Round the answers to at least four decimal places. 8. Big chickens: A report from a poultry industry news website stated that the weights of broilers (commercially raised chickens) are approximately normally distributed with mean 1450 grams and standard deviation 190 grams. Use the Cumulative Normal Distribution Table to answer the following. (a) Find the 23rd percentile of the weights. (b) Find the 89th percentile of the weights. (c) Find the first quartile of the weights. (d) A chicken farmer wants to provide a money-back guarantee that his broilers will weigh at least a certain amount. What weight should he guarantee so that he will have to give his customers' money back only 2% of the time? Round your answers to at least two decimals. 9. Tree heights: Cherry trees in a certain orchard have heights that are normally distributed with mean μ= 110 inches and standard deviation σ= 12 inches. Use the Cumulative Normal Distribution Table to answer the following. (a) Find the 24th percentile of the tree heights. (b) Find the 82nd percentile of the tree heights. (c) Find the third quartile of the tree heights. (d) An agricultural scientist wants to study the tallest 1% of the trees to determine whether they have a certain gene that allows them to grow taller. To do this, she needs to study all the trees above a certain height. What height is this? Round the answers to at least two decimal places .
Answered Same DayAug 02, 2021

Answer To: 1. Google it: According to a recent report, 68% of Internet searches in a particular month used the...

Pooja answered on Aug 03 2021
160 Votes
Sheet1
    1)
    a)
    P(good risk)
     =7773/11218
    0.6929
    2)
    P (V) = 0.47
    P(W) = 0.25
    P(V and
W) = .11
    a)
    P(V or W)
    P(V) + P(W) - P(VW)
     =0.47+0.25-0.11
    0.61
    b)
    P(V not)
    1-P(V)
     =1-0.47
    0.53
    5)
    Let A and B be events with P(A) = 0.3, P(B) = 0.4, and P(A and B) = 0.5. (a) Compute P(A and B) . (b) Are A and A mutually exclusive? Explain. (c) Are A and B independent? Explain.
    a)
    P(A and B) = 0.5
    b)
    two events are mutually exclusive when both cant occur at the same time
    here, P(AB) = 0.5
    Thus, A and B are not mutually exclusive
    c)
    for independent events, P(A)*P(B) = P(AB)
    But here, P(A)*P(B) = 0.3*0.4 = 0.12
    this is not equal to P(AB) =0.5
    this, A and B are not independent events
    6)
    a)
    P(A)
    ...
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