#37 M&Ms Data set 27 “M&M Weights” in Appendix B includes data from 100 M&M cand

Question

#37 M&Ms Data set 27 “M&M Weights” in Appendix B includes data from 100 M&M candies and 19 of them are green. Mars, Inc.claims that 16% of its plain M&M candies are green. For the following, assume that the claim of 16% is true, and assume that a sample consists of 100 M&Ms. a. Use the range rule of thumb to identify the limits separating values that are significantly low and those that are significantly high. Based on the results, is the result of 19 green M&Ms significantly high? b. Find the probability of exactly 19 green M&Ms. c. Find the probability of 19 or more green M&Ms. d. Which probability is relevant for determining whether the result of 19 green M&Ms is significantly high: the probability from part (b) or part (c)? Based on the relevant probability, is the result of 19 green M&Ms significantly high?

Explanation / Answer

Solutions:

a. Use the range rule of thumb to identify the limits separating values that are significantly low and those that are significantly high. Based on the results, is the result of 19 green M&Ms significantly high?

We are given

n = 100

p = 0.16 (16%)

q = 1 – p = 1 – 0.16 = 0.84

X = 19

Mean = n*p = 100*0.16 = 16

SD = sqrt(n*p*q) = sqrt(100*0.16*0.84) = 3.666061

According to range rule of thumb, range is four times the standard deviation.

So, range = 4*3.666061 = 14.66424

Now, according to thumb rule, most values lies with 2 standard deviations from mean.

Minimum limit = Mean - 2*SD = 16 - 2*3.666061 = 8.667878

Maximum limit = Mean + 2*SD = 16 + 2*3.666061 = 23.33212

Based on the results, the result of 19 green M&Ms is not significantly high. It would be significantly high if it is more than 23 M&Ms approximately.

b. Find the probability of exactly 19 green M&Ms.

We have to find P(X=19)

We are given

n = 100

p = 0.16 (16%)

q = 1 – p = 1 – 0.16 = 0.84

X = 19

Mean = n*p = 100*0.16 = 16

SD = sqrt(n*p*q) = sqrt(100*0.16*0.84) = 3.666061

By using continuity correction, we have to find P(18.5<X<19.5)

P(18.5<X<19.5) = P(X<19.5) – P(X<18.5)

For X < 19.5

Z = (X – mean)/SD

Z = (19.5 - 16) / 3.666061

Z = 0.954703

P(Z<0.954703) = P(X<19.5) = 0.830136

For X<18.5

Z = (18.5 - 16)/ 3.666061

Z = 0.681931

P(Z<0.681931) = P(X<18.5) = 0.752359

P(18.5<X<19.5) = P(X<19.5) – P(X<18.5) = 0.830136 - 0.752359 = 0.077777

P(X=19) = 0.077777

Required probability = 0.077777

c. Find the probability of 19 or more green M&Ms.

Here, we have to find P(X19)

By subtracting continuity correction 0.5, we have to find P(X18.5)

P(X18.5) = P(X>18.5) = 1 – P(X<18.5)

Z = (18.5 - 16)/ 3.666061

Z = 0.681931

P(Z<0.681931) = P(X<18.5) = 0.752359

P(X18.5) = P(X>18.5) = 1 – P(X<18.5) = 1 - 0.752359 = 0.247641

Required probability = 0.247641

d. Which probability is relevant for determining whether the result of 19 green M&Ms is significantly high: the probability from part (b) or part (c)? Based on the relevant probability, is the result of 19 green M&Ms significantly high?

The probability from part b is more significant because it indicate the very low probability of having 19 green M&Ms. This means result of 19 green M&Ms is significantly high. Based on the relevant probability, the result of 19 green M&Ms is significantly high.

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