Question

In: Statistics and Probability

Mr. Jackson could think about his experiment a little differently. How many times should he expect...

Mr. Jackson could think about his experiment a little differently. How many times should he expect to roll a 4 if he rolls his number cube 30 times? This is called the mean, or expected value. You can find it if you know the number of trials and the probability of success for an individual trial. Use the formula μx = np to find the expected value.

Part A: Find the expected value for the number of times you would roll a 4 in 30 rolls of a number cube.

a. How many trials, n, are there?

b. What is p, the probability of success for each trial?

c. What is the expected value of rolling a 4 in 30 rolls of a number cube?

Part B: You can also find the standard deviation for the binomial probability distribution of a specific outcome in a binomial experiment. Use the formula to find the standard deviation. You've already identified n and p in Part A. Show your work, and round your answer to two decimal places.

Part C: You've seen that a cumulative binomial probability is the probability of getting a number of successes within a given range. It's easiest to use your calculator's binomcdf command to find these. Find each of the probabilities described below. Show your work, and round your answer to three
decimal places.

a. What is the probability that Mr. Jackson will roll a 4 at most 5 times out of his 30 rolls of a
number cube?

b. What is the probability that Mr. Jackson will roll a 4 at least 5 times out of his 30 rolls of a number cube?

c. What is the probability that Mr. Jackson will roll a 4 between 3 and 7 times, inclusive, out of his 30 rolls of a number cube? Use this formula: P(a ≤ x ≤ b) = binomcdf(n,p,b) – binomcdf(n,p,a – 1).

Solutions

Expert Solution

A)

---------------

a)

Sample size , n =    30

b)
Probability of an event of interest, p =   0.1667

c)

Mean = np =    30   *   0.167   =           5.000

B)

-------------------------------

Standard deviation = √(np(1-p)) =   √   4.1667   =               2.04

C)

-------------------
a)

X P(X)
P ( X = 0) = C (30,0) * 0.166666666666667^0 * ( 1 - 0.166666666666667)^30= 0 0.0042
P ( X = 1) = C (30,1) * 0.166666666666667^1 * ( 1 - 0.166666666666667)^29= 1 0.0253
P ( X = 2) = C (30,2) * 0.166666666666667^2 * ( 1 - 0.166666666666667)^28= 2 0.0733
P ( X = 3) = C (30,3) * 0.166666666666667^3 * ( 1 - 0.166666666666667)^27= 3 0.1368
P ( X = 4) = C (30,4) * 0.166666666666667^4 * ( 1 - 0.166666666666667)^26= 4 0.1847
P ( X = 5) = C (30,5) * 0.166666666666667^5 * ( 1 - 0.166666666666667)^25= 5 0.1921

P(x<=5) = 0.6164

b)

X P(X)
P ( X = 0) = C (30,0) * 0.166666666666667^0 * ( 1 - 0.166666666666667)^30= 0 0.0042
P ( X = 1) = C (30,1) * 0.166666666666667^1 * ( 1 - 0.166666666666667)^29= 1 0.0253
P ( X = 2) = C (30,2) * 0.166666666666667^2 * ( 1 - 0.166666666666667)^28= 2 0.0733
P ( X = 3) = C (30,3) * 0.166666666666667^3 * ( 1 - 0.166666666666667)^27= 3 0.1368
P ( X = 4) = C (30,4) * 0.166666666666667^4 * ( 1 - 0.166666666666667)^26= 4 0.1847

P(x>=5) = 1- P(x<=4)

= 1- 0.4243

=0.5757

c)

X P(X)
P ( X = 3) = C (30,3) * 0.166666666666667^3 * ( 1 - 0.166666666666667)^27= 3 0.1368
P ( X = 4) = C (30,4) * 0.166666666666667^4 * ( 1 - 0.166666666666667)^26= 4 0.1847
P ( X = 5) = C (30,5) * 0.166666666666667^5 * ( 1 - 0.166666666666667)^25= 5 0.1921
P ( X = 6) = C (30,6) * 0.166666666666667^6 * ( 1 - 0.166666666666667)^24= 6 0.1601
P ( X = 7) = C (30,7) * 0.166666666666667^7 * ( 1 - 0.166666666666667)^23= 7 0.1098

P = 0.7835

Thanks in advance!

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