Respuesta :
Answer:
a) 0.25
b) 0.536
c) 0.25
d) 91
Step-by-step explanation:
We are given the following in the question:
X: 0 1 2
P(x) : 0.80 0.15 0.05
a) The mean of X
[tex]E(x) = \displaystyle\sum x_iP(x_i)\\E(x) = 0(0.80)+ 1(0.15) + 2(0.05)\\E(x) = 0.25[/tex]
b) The standard deviation of X
[tex]E(x^2) = \displaystyle\sum x_i^2P(x_i)\\E(x) = 0^2(0.80)+ 1^2(0.15) + 2^2(0.05)\\E(x^2) = 0.35\\\sigma^2 = E(x^2)-(E(x))^2\\\sigma^2 =0.35-(0.25)^2\\\sigma^2 = 0.2875\\\sigma = 0.536[/tex]
c) breakdowns are there per day
On average, there are 0.25 breakdowns per day.
d) breakdowns are expected during a one-year period
[tex]\text{Average breakdown per day}\times \text{Number of days}\\=0.25\times 365\\=91.25\\\approx 91[/tex]
Thus, there are approximately 91 breakdowns during a one-year period.
Probability distribution is the collection of value and probability pair for the given random variable. The figures for the given distribution is:
- Mean of X = 0.25
- Standard deviation of X ≈ 0.592
- On average, the number of breakdowns per day = 0.25
- Expected number of breakdowns in 365 working day year is ≈ 91
How to find the mean and variance of a random variable?
Supposing that the considered random variable is discrete, we get:
- Mean = [tex]E(X) = \sum_{\forall x_i} f(x_i)x_i\\\\[/tex]
- Variance = [tex]Var(X) = \sum_{\forall x_i} f(x_i)x^2_i\\[/tex]
As standard deviation is positive root of variance, thus,
[tex]\sigma = \sqrt {Var(X)}[/tex]
Let we suppose a random variable X tracking the number of breakdowns per day. Then,
For the given case, we have:
[tex]\begin{array}{cc}X = x_i&P(X = x_i) = f(x_i) \\0&0.80\\1&0.15\\2&0.05\end{array}[/tex]
Thus, we get:
[tex]E(X) = \sum_{\forall x_i} f(x_i)x_i = (0 \times 0.8 + 1 \times 0.15 + 2 \times 0.05)= 0.25[/tex]
Thus, mean of X = 0.30
And
[tex]Var(X) = \sum_{\forall x_i} f(x_i)x^2_i\\\\\\Var(X) = (0^2 \times 0.80 + 1^2 \times 0.15 + 2^2 \times 0.05) = 0.35[/tex]
Thus, standard deviation of X = [tex]\sqrt{0.35} \approx 0.592[/tex]
As X is tracking the number of breakdowns per day, thus, the average number of breakdowns per day is the mean of X which is 0.25
For getting average number of breakdown in a non-leap year, having work all days, we get:
Average number of breakdown in a 365 day working non-leap year = Average breakdowns per day × 365 = 0.25 × 365 = 91.25 ≈ 91
Thus, he figures for the given distribution is:
- Mean of X = 0.25
- Standard deviation of X ≈ 0.592
- On average, the number of breakdowns per day = 0.25
- Expected number of breakdowns in 365 working day year is ≈ 91
Learn more about expectation of a random variable here:
https://brainly.com/question/4515179
