A balanced die with 10 sides, numbered 1 through 10, is rolled 120 times. a. For the binomial distribution of Upper X equals 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 xequals0, would you be skeptical that the die is balanced? Explain why, based on the mean and standard deviation of X. d. Find the probability that xequals0.

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 , where is the cummulative distribution function of a standard normla random variable. The values of 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

Which is practically 0.

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.

19

step-by-step explanation: