Question 1178326: After watching a number of children playing games at a video arcade, a statistician estimated the following probability distribution of X, the number of games per visit.
X 1 2 3 4 5 6 7
P(x) .05 .15 .15 .25 .20 .10 .10
a. Determine the legitimacy of the distribution of the number of games per visits.
b. What is the probability that a child will play more than four games per visits?
c. What is the probability that a child will play at least two games per visits?
d. What is the expected number of games per visits?
e. What is the variance and standard deviation of the distribution?
f. Compute the coefficient of variation for the distribution
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Let's analyze this probability distribution step-by-step.
**a. Legitimacy of the Distribution**
For a probability distribution to be legitimate, two conditions must be met:
1. **All probabilities must be between 0 and 1 (inclusive).**
2. **The sum of all probabilities must equal 1.**
Let's check:
* All probabilities are between 0 and 1.
* Sum of probabilities: 0.05 + 0.15 + 0.15 + 0.25 + 0.20 + 0.10 + 0.10 = 1.00
Therefore, the distribution is legitimate.
**b. Probability of More Than Four Games**
We want to find P(X > 4), which is P(X = 5) + P(X = 6) + P(X = 7).
* P(X > 4) = 0.20 + 0.10 + 0.10 = 0.40
**c. Probability of At Least Two Games**
We want to find P(X ≥ 2), which is 1 - P(X < 2) = 1 - P(X = 1).
* P(X ≥ 2) = 1 - 0.05 = 0.95
**d. Expected Number of Games (Mean)**
The expected value (E[X]) is the mean of the distribution:
* E[X] = Σ[x * P(x)]
* E[X] = (1 * 0.05) + (2 * 0.15) + (3 * 0.15) + (4 * 0.25) + (5 * 0.20) + (6 * 0.10) + (7 * 0.10)
* E[X] = 0.05 + 0.30 + 0.45 + 1.00 + 1.00 + 0.60 + 0.70 = 4.10
**e. Variance and Standard Deviation**
* **Variance (Var[X]):**
* Var[X] = E[X²] - (E[X])²
* E[X²] = Σ[x² * P(x)]
* E[X²] = (1² * 0.05) + (2² * 0.15) + (3² * 0.15) + (4² * 0.25) + (5² * 0.20) + (6² * 0.10) + (7² * 0.10)
* E[X²] = 0.05 + 0.60 + 1.35 + 4.00 + 5.00 + 3.60 + 4.90 = 19.50
* Var[X] = 19.50 - (4.10)² = 19.50 - 16.81 = 2.69
* **Standard Deviation (σ):**
* σ = √Var[X] = √2.69 ≈ 1.6401
**f. Coefficient of Variation (CV)**
The coefficient of variation (CV) is the ratio of the standard deviation to the mean, expressed as a percentage:
* CV = (σ / E[X]) * 100
* CV = (1.6401 / 4.10) * 100 ≈ 40.00%
**Results**
* **a. Legitimacy:** The distribution is legitimate.
* **b. P(X > 4):** 0.40
* **c. P(X ≥ 2):** 0.95
* **d. E[X]:** 4.10 games
* **e. Var[X]:** 2.69, σ: 1.6401
* **f. CV:** 40.00%
|
|
|
