Question

Bag	Blue	Orange	Green	Yellow	Red	Brown	Total Number of Candies
1	9	13	14	10	7	7	60
2	13	10	6	9	9	8	55
3	13	12	4	10	9	6	54
4	16	13	8	6	6	8	57
5	10	10	12	5	15	4	56
6	9	18	3	6	12	12	60
7	11	13	6	15	8	6	59
8	12	18	5	9	6	5	55
9	12	10	8	15	10	1	56
10	12	14	8	12	7	9	62
11	11	10	9	11	8	8	57
12	13	11	11	7	10	5	57
13	10	15	7	10	8	8	58
14	13	14	6	8	8	5	54
15	10	14	5	6	13	10	58
16	15	14	5	8	5	10	57
17	11	16	9	8	5	9	58
18	14	9	8	7	9	6	53
19	13	7	8	8	9	8	53
20	10	10	7	10	11	11	59
21	8	7	7	13	14	10	59
22	8	11	9	10	5	11	54
23	7	8	9	8	8	13	53
24	10	11	6	7	12	10	56
25	9	13	7	8	9	8	54
26	10	6	13	9	9	8	55
27	10	12	9	8	10	7	56
28	11	12	7	12	10	8	60
29	9	14	7	9	8	6	53
30	10	10	7	10	11	11	59

We evaluated the number and distribution of M&M Minis® candies. A sample of 30 bags was collected and the data is reported in the attached Excel file. This study examined the color percentages and the total number of candies in each bag.

If a random sample of 50 bags is selected, find the probability that the mean number of candies in a bag is at least 56 candies. Assume the total number of candies is normally distributed. F) Assume the total number of candies is normally distributed, find a 95% confidence interval for the mean total number of candies.

Answer 1

Let X denotes the total number of candies in a bag.

Since the distribution of X is normal.

X ~ N ( mu, sigma²).

Xbar = Mean number of candies in bag.

The distribution of Xbar is approximately normal.

Xbar ~ N (mu, sigma²/n).

Since Xbar = sum(Xi) / n and S2 = sum(Xi- Xbar)^2 / n-1 is an unbiased estimators of mu and sigma² respectively.

Total Number of Candies (X)	(Xi- Xbar)^2
60	11.7875
55	2.4545
54	6.5879
57	0.1877
56	0.3211
60	11.7875
59	5.9209
55	2.4545
56	0.3211
62	29.5207
57	0.1877
57	0.1877
58	2.0543
54	6.5879
58	2.0543
57	0.1877
58	2.0543
53	12.7213
53	12.7213
59	5.9209
59	5.9209
54	6.5879
53	12.7213
56	0.3211
54	6.5879
55	2.4545
56	0.3211
60	11.7875
53	12.7213
59	5.9209
1697	181.3667

S = sqrt(6.2540) = 2.5008

n = number of bags randomly selected = 50.

P ( Mean number of candies in a bag at least 56 candies) = P ( Xbar >= 56)

since

= P ( Z >= -1.6023) = 0.9455

P ( Mean number of candies in a bag at least 56 candies) = 0.9455.

Alpha: level of significance = 0.05.

Z_alpha/2 = Z_0.025 = 1.96

95% confidence interval for mean total number of candies is

( Xbar - (S/sqrt(n) * Z_alpha/2, Xbar + (S/sqrt(n) * Z_alpha/2)

= ( 56.5667 - ( 2.5008/ sqrt(50) * 1.96 , 56.5667 - ( 2.5008/ sqrt(50) * 1.96)

= ( 55.8735, 57.2599)