Question

To increase sales, a local donut shop began putting an extra donut in some of the boxes. Customers are unaware of which boxes had the extra donut and the shop owner claimed that one in seven boxes had an extra donut. Eight customers each bought one box of donuts at the shop. Let X = the number of customers that bought a box containing an extra donut.
Part A: Is X a binomial random variable? Explain.

Part B: What is the mean and standard deviation of X? Provide an interpretation for each value in context.

Part C: Two of the eight customers buy a box of donuts with an extra donut. Is the shop's claim accurate? Compute P(X ≥ 2) and use the result to justify your answer.

Answer 1

Answer:

Step-by-step explanation:

Hello!

Part A

First, determine your study variable:

X: Number of boxes with an extra donut in a sample of eight.

To see if the variable has a Binomial distribution you have to check if the binomial criteria are met:

1. The number of observation of the trial is fixed (In this case n = 8, the boxes each customer bought make the sample)

2. Each observation in the trial is independent, this means that none of the trials will affect the probability of the next trial (In this case, the amount of donuts in one box does not affect on the probability of the next box having an extra donut)

3. The probability of success in the same from one trial to another (In this case our "success" will be that the box has an extra donut, according to the owners claim that is 1/7; ρ=0,14)

So X≈ Bi (n;ρ)

Where n represents the sample (n=8) and ρ is the probability of success (ρ=0.14)

Part B

The mean of the binomial distribution is E(X)= nρ

E(X)= 8 * 0.14= 1.12

The mean of the distribution is also called the expected value. You'd expect that 1.12 boxes have an extra donut.

The variance of the binomial distribution is V(X)= nρ(1 - ρ)

V(X)= 8*0.14*(1 - 0.14)= 0.9632

Its square root is the standard deviation

√V(X)= 0.98

The standard deviation is a measure of dispersion, it indicates how much the values ​​of the distribution of the central value are separated. In this case, 0.98 indicates that the distribution of the number of boxes with an extra donut is far from the expected value.

Part C

Two of the eight customers buy a box with an extra donut, symbolically:

P(X=2) = P(X≤2) - P(X≤1)= 0.91 - 0.68 = 0.22

There is a 22% chance that two customers bought a box with an extra donut.

Compute:

P(X≥2)= 1 - P(X<2)= 1 - P(X≤1)= 1 - 0.68= 0.32

There is a 32% chance that two or more customers bought a box with an extra donnut.

I hope it helps!