As the M&M inspector, you are charged with certifying that the size of the M&M is consistent. You open a package of M&M's and find the following weights of the M&M's measured in grams:

0.884 0.882 0.885 0.886 0.880 0.884 0.883 0.881 0.883 0.882

0.885 0.884 0.887 0.889 0.882 0.884 0.885 0.883 0.884 0.888

0.879 0.881 0.887 0.877 0.889 0.89 0.884 0.885 0.883 0.882

Mean, Median, Mode, Maximum, Minimum, Range, and Standard Deviation for this package of M&M's.

• Mean = 0.8839
• Median = (0.884+0.884)/2 = 0.884 ( there are 30 values so the median is the average of the 15th and the 16th)
• Mode= it is the number that appears most frequently in a set = 0.884
• Maximum = 0.890
• Minimum = 0.877
• Range= Max - min = 0.890 -0.877 = 0.013
• Standard Deviation = using Excel I got 0.0029

Based on the mean and standard deviation of your pack of M&M's, and assuming that the size of M&M's is normally distributed, here is the range of weights of the middle 68.3%, 95.4%, and 99.7% of M&M's.

• the middle 68.3% is 1 = 0.884 0.003 so [0.881 ;0.887] its range is 2 = 0.006
• the middle 95.4% is 2 = 0.884 0.006 so [0.878 ;0.890] its range is 4 = 0.012
• the middle 99.7% is 3 = 0.884 0.009 so [0.875 ;0.893] its range is 6 = 0.018

The probability that an M&M will weigh more than 0.883 grams

Standard Deviation, : 0.0029

and Mean, :0.88422857142857

So Z-score of .883 gram M&M is

(.883-.884)/.0029

=-0.344827586

P-value from Z-Table

Probability of x>.883= 0.63716

P(M&M size >.883)=.63716

The probability that an M&M will weigh less than 0.885 grams.

• = 0.884 and = 0.0029 so
• for X = 0.885 , z =(0.885-0.884)/0.0029 = 0.3448
• P( X <0.885) =P(z<0.3448) = 0.63488 (from Z tables)

Using the z-scores for M&M's weighing 0.883 grams and 0.885 grams, the probability an M&M randomly chosen from the package will weigh between 0.883 grams and 0.885 grams.

So = 0.884 and = 0.0029

• for X = 0.885 , z =(0.885-0.884)/0.0029 = 0.3448
• P( X <0.885) =P(z<0.3448) = 0.63488 (from Z tables)
• for X = 0.883 , z =(0.883-0.884)/0.0029 = - 0.3448
• P( 0.883

If the package contained 5 Red, 5 Orange, 4 Green, 6 Blue, 3 Yellow, and 7 Brown M&M's, the probabilities of pouring three M&M's from the package and all three being the same color. The probability of all three being a different color.

• P(Red) = 5/30 = 1/6 , P(Orange) = 1/6 , P(Green) = 4/30 = 2/15 , P(Blue) = 6/30 = 1/5 , P(R=Yellow) = 3/30 = 1/10 , P(Brown) = 7/30

• the probabilities of pouring three M&M's from the package and all three being the same color = [C(5,3)+C(5,3)+C(4,3)+C(6,3)+C(3,3)+C(7,3)]/C(30,3) = ( 10+10+4+20+1+35) /4060 = 0.0197
• the probability all three are a different color: P = 1 - P( 3 same color) - P( 2 same color) =
• P( 2 same color) = [C(5,2)*C(25,1)+C(5,2)*C(25,1)+C(4,2)*C(26,1)+C(6,2)*(24,1)+C(3,2)*(27,1)+C(7,2)*C(23,1)]/C(30,3) = (10*25+10*25+4*26+20*24+1*27+35*23) /4060 = 1916/4060 = 0.47192
• Thus P = 1 - 0.0197 - 0.47192 = 0.5084

Make a report to give your new employer containing a table of M&M statistics, and the findings. Be sure to include your professional opinion about the consistency of weight and variety of color.