Question 1 What is the sample size, n? ( i think this is 10) Question 2 What is
ID: 3247011 • Letter: Q
Question
Question 1
What is the sample size, n? ( i think this is 10)
Question 2
What is probability of a success, ? Give in the form of a decimal.
Question 3
What is the average number of orange M&M’s? ( i think this is 2)
Question 4
What is the standard deviation of orange M&M’s? Give solution to three decimal places.
Question 5
Find the probability of selecting five or fewer orange M&M’s out of a sample of ten. Give solution to four decimal places.
Question 6
When sampling 10 M&M’s, what is the probability of getting exactly four that are orange? Give solution to four decimal places.
Question 7
What is the probability that the number of orange M&M’s drawn out of ten is more than three, but less than nine? Give solution to four decimal places.
Question 8
Find the probability of selecting more than 2 orange M&M’s out of a sample of ten. Give solution to four decimal places.
Question 9
When sampling 10 M&M’s, what is the probability of getting five that are not orange? Give solution to four decimal places.
Color red orange yellow green blue brown total Avg Class Estimate 7 15 7 9 14 7 59 Percentage based on Class Estimate 12% 25% 12% 15% 24% 12% 100% Per M&M;'s Official 13% 20% 14% 16% 24% 13% 100%Explanation / Answer
Answer to question# 1)
The sample size in this case is total of the average class estimate
We have total = 59
thus the sample size is 59
.
Answer to question# 2)
Probability of success depends on how is success defined. Suppose if success is defined as color "red"
Then Probability of success = 0.12 ( that is the 12% that is mentioned in the third column of the table above)
.
Answer to question# 3)
Average number of oranges M & M's is
20% of 59 = 0.2 *59 = 11.8 = 12
.
Answer to question# 4)
Since p = 0.2 , n = 59 ........[for M & M's]
Standard deviation = sqrt(n * p *(1-p))
Standard deviation = sqrt(59 * 0.2 * 0.8)
Standard deviation = sqrt(9.44)
Standard deviation = 3.0725
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