To increase sales, an online clothing store began giving a 50% off coupon to random customers. Customers didn't know whether they would receive the coupon until after the final sale. The website claimed that one in five customers received the coupon. Six customers each made purchases from the website. Let X = the number of customers that received the 50% off coupon. 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 six customers receive the coupon with their purchase. Is the store's claim accurate? Compute P(X ≥ 2) and use the result to justify your answer.

Yes, X is a Binomial Distribution. Mean & Standard Deviation are 1.2, 0.9798  respectively. The company's claim is correct.

A] Binomial Distribution finds probability of 'r' successes out of 'n' trials, where probability of success and failure are 'p' & 'q' respectively.

Here : Success is to get discount coupon with probability 'p' = 1/5 = 20% = 0.2, hence failure probability of not getting discount coupon 'q' = 1 - 0.2 = 0.8 ,       n trials are like number of customers = 10, 'r' implies number of customers who get discount

B] Mean, ie E (X) = np =  6 x 0.2 = 1.2

Standard deviation = √ ( np ) ( 1 - p)  

= √ 1.2 (0.8) = √ 0.96 = 0.9798

C] n = 6 & prob (r > 2)

Prob (r = k) = N c r .

P ( r > 2 ) = 1 - P ( r < 2 )  = 1 - [ P (r = 0) + P (r = 1) ]

= 1 - [ 6c1 (0.2)^6 (0.8)^0 + 6c2 (0.2)^1 (0.8)^5 ]

= 1 - 0.6554 = 0.3446

As the probability of getting 3 or more discount coupons from 6 customers is 34.4%, which is higher than company claim of 20% customers getting discount coupon. So, company's claim is correct.

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a. It is a random binomial variable

b. mean = 1.2 and standard deviation = 0.9798

c. The store's claim is inaccurate

Step-by-step explanation:

PART A

Yes, X is a random binomial variable, with a probability of 1/5 = 0.2

Each customer can be seen as a Bernoulli variable with P = 0.2 , and their sum is binomial random variable.

PART B

Here, we want to calculate the mean and standard deviation of X.

n = 6 and P = 0.2

Mathematically, mean = np = 6 × 0.2 = 1.2

Standard deviation = =

PART C

We proceed using the binomial approximation here;

P (X ≥ 2) = 1- P(X≤1) = 1 - P(X=0) - P(X=1)

= 6C0 (0.8)^6 -6C1(0.2)(0.8)^5 = 1-0.262144-0.393216 = 0.34464

What the above means is that there are 34% chances that more than or equal to 2 customers may receive the coupon out of the six customers.

Hence, the claim of the company is not accurate. This is because the claim  of the company is 1/6 = 33.33% and our computation is clearly more than this.

x =±sqrt(1/3)

step-by-step explanation:

x+4/6x=1/x

multiply each side by 6x to get rid of the fractions

6x*(x+4/6x)=1/x*6x

distribute

6x^2 +4 = 6

subtract 4 from each side

6x^2 +4-4 = 6-4

6x^2 =2

divide by 6

6x^2/6 = 2/6

x^2 = 1/3

take the square root of each side

sqrt(x^2) = ±sqrt(1/3)

x =±sqrt(1/3)