To solve this problem, we can model it using a binomial distribution. A binomial distribution arises when we have a fixed number of independent trials (e.g., trucks under maintenance), each with two possible outcomes (e.g., delivered on time or not delivered on time), and the probability of success (e.g., delivered on time) is the same for each trial.
Here, we have:
- \( n = 6 \) (the number of trucks)
- \( p = 0.60 \) (the probability of a truck being delivered on time)
- \( q = 1 - p = 0.40 \) (the probability of a truck not being delivered on time)
### (a) Constructing the Probability Distribution
The probability of exactly \( k \) trucks being delivered on time out of \( n \) trucks can be calculated using the binomial probability formula:
\[ P(X = k) = \binom{n}{k} p^k q^{n-k} \]
Where \( \binom{n}{k} \) represents the binomial coefficient:
\[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]
Let's calculate \( P(X = k) \) for \( k = 0 \) to \( k = 6 \):
1. \( P(X = 0) \):
\[ P(X = 0) = \binom{6}{0} (0.60)^0 (0.40)^6 = 1 \times 1 \times 0.40^6 \approx 0.0041 \]
2. \( P(X = 1) \):
\[ P(X = 1) = \binom{6}{1} (0.60)^1 (0.40)^5 = 6 \times 0.60 \times 0.40^5 \approx 0.0369 \]
3. \( P(X = 2) \):
\[ P(X = 2) = \binom{6}{2} (0.60)^2 (0.40)^4 = 15 \times 0.36 \times 0.40^4 \approx 0.1382 \]
4. \( P(X = 3) \):
\[ P(X = 3) = \binom{6}{3} (0.60)^3 (0.40)^3 = 20 \times 0.216 \times 0.40^3 \approx 0.2765 \]
5. \( P(X = 4) \):
\[ P(X = 4) = \binom{6}{4} (0.60)^4 (0.40)^2 = 15 \times 0.1296 \times 0.40^2 \approx 0.3110 \]
6. \( P(X = 5) \):
\[ P(X = 5) = \binom{6}{5} (0.60)^5 (0.40)^1 = 6 \times 0.07776 \times 0.40 \approx 0.1866 \]
7. \( P(X = 6) \):
\[ P(X = 6) = \binom{6}{6} (0.60)^6 (0.40)^0 = 1 \times 0.07776 \approx 0.0778 \]
So the probability distribution is:
| \( k \) | \( P(X = k) \) |
|---------|-----------------|
| 0 | 0.0041 |
| 1 | 0.0369 |
| 2 | 0.1382 |
| 3 | 0.2765 |
| 4 | 0.3110 |
| 5 | 0.1866 |
| 6 | 0.0778 |
### (b) Expected Value and Standard Deviation
The expected value \( E(X) \) for a binomial distribution is given by:
\[ E(X) = np = 6 \times 0.60 = 3.6 \]
The variance \( \text{Var}(X) \) for a binomial distribution is given by:
\[ \text{Var}(X) = npq = 6 \times 0.60 \times 0.40 = 1.44 \]
The standard deviation \( \sigma(X) \) is the square root of the variance:
\[ \sigma(X) = \sqrt{\text{Var}(X)} = \sqrt{1.44} = 1.2 \]
So:
- The expected value (mean) of the distribution is \( 3.6 \) trucks.
- The standard deviation of the distribution is \( 1.2 \) trucks.
truck operator has determined that a car repair shop derivers maintained trucks on schedule 60% of the time. If the operator has 6 trucks under maintenance (a) construct the probability distribution for the number of truck to be delivered on time. (b) Find the expected value and the standard deviation of the distribution.
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