Question

A balanced die with six sides is rolled 60 times. a. For the binomial distribution of X = number of 6s, what is n and what is p? b. Find the mean and the standard deviation of the distribution of X. Interpret. c. If you observe x = 0, would you be skeptical that the die is balanced? Explain why

Answer 1

a) n=60 , p=1/6

b) mean= 10 , std = 2.88

c) The dice is probably not balanced

Step-by-step explanation:

In these experiment , the random variable X= number of 6s has a binomial distribution with

n = number of independent rolls of the dice = 60

p = probability to get a 6 for every independent roll of the dice = 1/6 (since the dice is balanced)

For a binomial distribution , the expected value is

E(X) = n*p = 1/6*60 = 10

and the standard deviation is

σ(X) =√( n*p*(1-p)) = √(60*1/6*5/6)= 2.88

finally since the binomial probability is :

P(X=x)=n!/[(n-x)!*x!]*p^x*(1-p)^(n-x)

then for x=0 :

P(X=0)= n!/[(n-0)!*0!]*p^0*(1-p)^(n-0) = (1-p)^n = (5/6)^60 = 1.77*10⁻⁵

therefore is really unlikely to not observe a single 6 in 60 rolls , and therefore we could suspect that the dice is not really balanced

Answer 2

a) n = 60, p = 1/6

b)
`\mu = 10` , SD = 2.89

c) I will be skeptical that the die is balanced

Step-by-step explanation:

a) X = number of 6s

The total number of times the die is rolled is 60, therefore, n = 60

Since the die is balanced, probability that a 6 is obtained,

`p = (number of possible outcomes)/(total number of outcomes)`

`p = (1)/(6)`

b1) For a binomial distribution, mean,
`\mu = np`

`\mu = 60 *(1)/(6) \\\mu = 10`

Standard Deviation,
`SD = √(npq)`

`q = 1-p\\q = 1- (1)/(6) \\q = (5)/(6)`

`SD = \sqrt{60*(1)/(6) *(5)/(6) }`

SD = 2.89

c) If observing x = 0 will still mean that the die is balanced, then the mean must have its boundary within 3 standard deviation.

`\mu - 3 SD = 10 - 3(2.89) = 1.33\\\mu + 3 SD = 10 + 3(2.89) = 18.67`

x = 0 is above 3 standard deviations from the mean, if I observe x = 0, I would be skeptical that the die is balanced