Respuesta :
Answer:
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:
a) n = 60, p = 1/6
b) [tex]\mu = 10[/tex] , SD = 2.89
c) I will be skeptical that the die is balanced
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,
[tex]p = \frac{number of possible outcomes}{total number of outcomes}[/tex]
[tex]p = \frac{1}{6}[/tex]
b1) For a binomial distribution, mean, [tex]\mu = np[/tex]
[tex]\mu = 60 *\frac{1}{6} \\\mu = 10[/tex]
Standard Deviation, [tex]SD = \sqrt{npq}[/tex]
[tex]q = 1-p\\q = 1- \frac{1}{6} \\q = \frac{5}{6}[/tex]
[tex]SD = \sqrt{60*\frac{1}{6} *\frac{5}{6} }[/tex]
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.
[tex]\mu - 3 SD = 10 - 3(2.89) = 1.33\\\mu + 3 SD = 10 + 3(2.89) = 18.67[/tex]
x = 0 is above 3 standard deviations from the mean, if I observe x = 0, I would be skeptical that the die is balanced
