A candy company claims that 16% of its plain candies are orange, and a sample of
ID: 3127495 • Letter: A
Question
A candy company claims that 16% of its plain candies are orange, and a sample of 100 such candies is randomly selected. a. Find the mean and standard deviation for the number of orange candies in such groups of 100. mu equals= nothing sigma equals= nothing (Round to one decimal place as needed.) b. A random sample of 100 candies contains 26 orange candies. Is this result unusual? Does it seem that the claimed rate of 16% is wrong? A. Yes comma because 26 is greater than the maximum usual value.Yes, because 26 is greater than the maximum usual value. Thus, the claimed rate of 16 % is probably wrong. B. Yes comma because 26 is below the minimum usual value.Yes, because 26 is below the minimum usual value. Thus, the claimed rate of 16 % is not necessarily wrong. C. No comma because 26 is within the range of usual values.No, because 26 is within the range of usual values. Thus, the claimed rate of 16 % is probablyprobably wrong. D. Yes, because 26 is within the range of usual values. Thus, the claimed rate of 16% is probably wrong.
Explanation / Answer
a.
Mean ( np ) =100 * 0.16 = 16
Standard Deviation ( npq )= 100*0.16*0.84 = 3.6661 ~ 3.7
Normal Distribution = Z= X- u / sd
b.
A random sample of 100 candies contains 26 orange candies
P(X > 26) = (26-16)/3.6661
= 10/3.6661 = 2.7277
= P ( Z >2.728) From Standard Normal Table
= 0.0032
A. Yes comma because 26 is greater than the maximum usual value.Yes, because 26 is greater than the maximum usual value. Thus, the claimed rate of 16 % is probably wrong.
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