Question

1. As a special promotion for its 12-ounce cans of cold coffee, a coffee drink company printed a message on the bottom of each can. Some of the bottoms read, "Better luck next time!" whereas others read, "You won!" The company advertised the promotion with the slogan "One in five wins a prize!" Suppose the company is telling the truth and that every 12-ounce can of coffee has a one-in-five chance of being a winner. Six friends each buy one 12-ounce can of coffee at a local convenience store. Let X equal the number of friends who win a prize.

Part A: Explain why X is a binomial random variable. (3 points)

Part B: Find the mean and standard deviation of X. Interpret each value in context. (4 points)

Part C: The store clerk is surprised when two of the friends win a prize. Is this group of friends just lucky, or is the company's one-in-five claim inaccurate? Compute P(X ≥ 2) and use the result to justify your answer.

2. The army reports that the distribution of waist sizes among female soldiers is approximately normal, with a mean of 28.4 inches and a standard deviation of 1.2 inches.

Part A: A female soldier whose waist is 26.1 inches is at what percentile? Mathematically explain your reasoning and justify your work. (5 points)

Part B: The army uniform supplier regularly stocks uniform pants between sizes 24 and 32. Anyone with a waist circumference outside that interval requires a customized order. Describe what this interval looks like if displayed visually. What percent of female soldiers requires custom uniform pants? Show your work and mathematically justify your reasoning.

Answer 1

(1)

Part A: Explain why X is a binomial random variable. (3 points)

The printed message only one of the two messages : better luck next time or You won. Therefore there are only two possible outcomes. Also the probability of winning is same for all coffee : one in five. Hence, it is a binomial distribution since there are only two possible outcomes and each can has same probability.

Part B: Find the mean and standard deviation of X. Interpret each value in context. (4 points)

There are six friends so

E(X) = np

On average out of six chances(friends) 1.2 are expected winning chances.

SD =

=

On average 0.98 chances away from the expected winning chance.

Part C: The store clerk is surprised when two of the friends win a prize. Is this group of friends just lucky, or is the company's one-in-five claim inaccurate? Compute P(X ≥ 2) and use the result to justify your answer.

In binomial

in our case

P(X ≥ 2) = 1 - P( X <2)

= 1 - [P(X = 0) +P(X = 1) ]

= 1 - (0.2621 + 0.3932)

Two friends out of six won that is their probability of winning = 2 / 6 = 0.3333

This seems more of less consistent with the company's claim, therefore we can that the friends got lucky.

(2)

Part A: A female soldier whose waist is 26.1 inches is at what percentile? Mathematically explain your reasoning and justify your work. (5 points)

Percentile is the value below which the percentile (p % ) of population lies.

So a soldier with waist 26.1 is at the following percentile

= P(Z < -1.92

= 1 - P( Z < 1.92)

= 1 - 0.9724

= 0.0276

  

This is 3%.

26.1 waist lies at the 3rd percentile.

Part B: The army uniform supplier regularly stocks uniform pants between sizes 24 and 32. Anyone with a waist circumference outside that interval requires a customized order. Describe what this interval looks like if displayed visually. What percent of female soldiers requires custom uniform pants? Show your work and mathematically justify your reasoning.

The interval looks like

24 < X <32 : The waist size being being within the range of 24 to 32.

Percent of the females who require customized pants can be calculated by subtracting P(24 < X < 32) from 1

Percent of female required customized pants = 1 - P( 24 < X < 32)

P(24 < X < 32) = P( X < 32) - P( X < 24)

= P( Z < 3 ) - P( Z < -3.67)

= P( Z < 3) - [1 - P( Z < 3.67) ]

= 0.9987 - (1 - 0.9999)

= 0.9985

Percent of female required customized pants = 1 - P( 24 < X < 32)

= 1 - 0.9985

= 0.0015

female soldiers require customized pants.