Question

Are 16% of M&M’s Green? Mars, Inc. claims that 16% of its M&M plain candies are green. A sample of 100 M&Ms is randomly selected. a. Find the mean and standard deviation for the numbers of green M&Ms in such groups of 100. b. Data Set 18 in Appendix B consists of a random sample of 100 M&Ms in which 19 are green. Is this result unusual? Does it seem that the claimed rate of 16% is wrong?

Answer 1

Answer: This result is not unusual.

So, it does not seem that the claimed rate of 16% is  wrong

Step-by-step explanation:

Since we have given that

n = 100

p = 0.16

q = 1 - 0.16 = 0.84

So, Mean would be

$np=100\times 0.16=16$

and standard deviation would be

$\sqrt{np(1-p)}=\sqrt{100\times 0.16\times 0.84}=3.67$

Now, if there are 19 green,

so, we will check how it is far,

$\dfrac{19-16}{3.67}=\dfrac{3}{3.67}=0.817$

Since it is 0.817 standard deviations away from the mean.

So, this result is not unusual.

So, it does not seem that the claimed rate of 16% is  wrong.