Respuesta :
Answer:
a) n=120, p = 0.1
b) The mean is μ=12, and the standard deviation is σ = 3.2863
c) Yes, the die is most likely not balanced
d) P(X=0) = 3.2292 e^{-6}
Step-by-step explanation:
a) Since we roll the die 120 times, then n = 120. p is the probability that a balanced die rolls 6, which is 1/10, because there are a total of 10 possible outcomes.
b) The mean of X is np = 120*0.1 = 12. The standard deviation of X is √(np(1-p)) = √(12*0.9) = √10.8 = 3.2863
c) Since the number of experiments is high enough (n>30), the central limit theorem states that X distribution can be approximated by a normal distribution with similar mean and standard deviation than X.
If we take the mean 12 and we move 3 standard deviations to the left we obtain 12-3*3.2863 = 2.141. Which is way at the right of 0. This means that 0 is more than 3 standard deviations at the left of the mean. The probability to be 3 standard deviations at the left of the mean in a random variable with normal distribution is [tex] \phi(-3) = 1-\phi(3) = 1- 0.9987 = 0.0013 [/tex] , where [tex] \phi [/tex] is the cummulative distribution function of a standard normla random variable. The values of [tex] \phi [/tex] can be found in the attached file.
This means that it is highly unlikely to take 0 as a result if the mean were to be 12, in other words, if the probability to obtain 6 is 1/10 (or more specifically, the dice is balanced). This shows evidence that the die is most likely not balanced.
d) Using the formula of a random variable we have
[tex]P(X=0) = {120 \choose 0} * 0.1^0 * (1-0.1)^{120} = 0.9^{120} = 3.2292 \, e^{-6}[/tex]
Which is practically 0.
