1MM plain candies have a mean weight of 08565 g and a standa

1.M&M plain candies have a mean weight of 0.8565 g and a standard deviation of 0.0518 g (based on Data Set 20 in Appendix B). The M&M candies used in Data Set 20 came from a package containing 465 candies, and the package label stated that the net weight is 396.9 g. a) If one M&M plain candy is randomly selected, find the probability that it weighs between 0.8535 g and 0.8565 g

c) If a package (containing 465 M&M plain candies) is randomly selected, find the probability its mean weight is between 0.8535 g and 0.8565 g.

2. The Mars company claims that 17% of its M&M candies are red. a) Using the Data Set 20 in Appendix B, find the sample proportion of M&Ms that are red. b) Construct a 90% confidence interval estimate of the proportion of red M&Ms.c) Does it appear that the claimed rate of 17% is correct?

Solution

1)

(a)

P(0.8535 < X < 0.8565) = P( (0.8535 - 0.8565)/0.0518 <z<(0.8565 - 0.8565)/0.0518))

= P( -0.06 < z < 0)

= 0.5 - 0.4761

= 0.0239 Answer

(c)

P(0.8535 < X < 0.8565) = P( (0.8535 - 0.8565)/(0.0518/sqrt(465)) <z<(0.8565 - 0.8565)/(0.0518/sqrt(465)))

= P( -1.25< z < 0)

= 0.5 -0.1056

= 0.3944 Answer

2)

P(Red candies) = 0.17

a)

Sample proportion of red candies = 0.17 * 465 = 79.05 ~ 79 Answer

b)

p =0.17

Margin of error = (1.645 * 0.0518) / sqrt(465) = 0.0039

Confidence Interval : (0.17 - 0.0039 , 0.17 + 0.0039)

= (0.1661 , 0.1739) Answer

(c)

Yes claim is correct since 0.17 lies in the confidence interval.