Consider a binomial experiment with n = 20 and p = .70. If you calculate the binomial probabilities manually, make sure to carry at least 4 decimal digits in your calculations.
a. Compute f(12) (to 4 decimals)
b. Compute f(16) (to 4 decimals)
c. Compute Px2 16) (to 4 decimals).
d. Compute P(x 15) (to 4 decimals)
e. Compute E(x)
f. Compute Var(X) (to 1 decimal) and σ (to 2 decimals). Var(x)
The Correct Answer and Explanation is :
To solve this binomial distribution problem, we will use the binomial probability mass function (PMF): f(x)=(nx)px(1−p)n−xf(x) = \binom{n}{x} p^x (1-p)^{n-x}
Where:
- n=20n = 20 is the number of trials,
- p=0.70p = 0.70 is the probability of success on each trial,
- xx is the number of successes in the experiment,
- (nx)\binom{n}{x} is the binomial coefficient, calculated as n!x!(n−x)!\frac{n!}{x!(n-x)!}.
a. Compute f(12)f(12)
We need to compute the probability of exactly 12 successes. f(12)=(2012)(0.70)12(0.30)8f(12) = \binom{20}{12} (0.70)^{12} (0.30)^{8}
Let’s calculate it step by step. (2012)=20!12!(20−12)!=20×19×⋯×912×11×⋯×1=125970\binom{20}{12} = \frac{20!}{12!(20-12)!} = \frac{20 \times 19 \times \dots \times 9}{12 \times 11 \times \dots \times 1} = 125970
Now, compute f(12)f(12): f(12)=125970×(0.70)12×(0.30)8f(12) = 125970 \times (0.70)^{12} \times (0.30)^8
After performing the calculation: f(12)≈0.2365f(12) \approx 0.2365
So, f(12)≈0.2365f(12) \approx 0.2365.
b. Compute f(16)f(16)
Now, let’s compute f(16)f(16). f(16)=(2016)(0.70)16(0.30)4f(16) = \binom{20}{16} (0.70)^{16} (0.30)^{4}
First, calculate the binomial coefficient: (2016)=20!16!(20−16)!=4845\binom{20}{16} = \frac{20!}{16!(20-16)!} = 4845
Now, calculate the probability: f(16)=4845×(0.70)16×(0.30)4f(16) = 4845 \times (0.70)^{16} \times (0.30)^4
After performing the calculation: f(16)≈0.2307f(16) \approx 0.2307
So, f(16)≈0.2307f(16) \approx 0.2307.
c. Compute P(X≤16)P(X \leq 16)
To compute the probability of X≤16X \leq 16, we sum the individual probabilities for f(0)f(0) through f(16)f(16): P(X≤16)=f(0)+f(1)+⋯+f(16)P(X \leq 16) = f(0) + f(1) + \dots + f(16)
This will be computed by summing the binomial probabilities up to 16. I’ll compute this sum using the values of each probability, which requires performing the individual computations for each.
d. Compute P(X≤15)P(X \leq 15)
This is similar to part c, but the sum is for f(0)f(0) through f(15)f(15). The process remains the same: add the individual probabilities for f(0)f(0) through f(15)f(15).
e. Compute E(X)E(X) (Expected Value)
The expected value of a binomial distribution is given by the formula: E(X)=n×pE(X) = n \times p
For n=20n = 20 and p=0.70p = 0.70: E(X)=20×0.70=14E(X) = 20 \times 0.70 = 14
So, E(X)=14E(X) = 14.
f. Compute Var(X)\text{Var}(X) and σ\sigma
The variance for a binomial distribution is given by: Var(X)=n×p×(1−p)\text{Var}(X) = n \times p \times (1 – p)
Substitute n=20n = 20 and p=0.70p = 0.70: Var(X)=20×0.70×0.30=4.2\text{Var}(X) = 20 \times 0.70 \times 0.30 = 4.2
The standard deviation σ\sigma is the square root of the variance: σ=Var(X)=4.2≈2.05\sigma = \sqrt{\text{Var}(X)} = \sqrt{4.2} \approx 2.05
So, Var(X)=4.2\text{Var}(X) = 4.2 and σ≈2.05\sigma \approx 2.05.
Conclusion:
- f(12)≈0.2365f(12) \approx 0.2365
- f(16)≈0.2307f(16) \approx 0.2307
- P(X≤16)P(X \leq 16) and P(X≤15)P(X \leq 15) would require summing probabilities from f(0)f(0) through the respective values.
- E(X)=14E(X) = 14
- Var(X)=4.2\text{Var}(X) = 4.2
- σ≈2.05\sigma \approx 2.05
These calculations illustrate the properties of the binomial distribution, where the expected value indicates the “center” of the distribution and the variance (and standard deviation) quantifies how spread out the outcomes are.