Respuesta :
a) Add up all the probabilities [tex]f(x)[/tex] where [tex]x>3[/tex]:
[tex]f(4)+f(5)+f(6)+f(7)=0.35[/tex]
b) The expected value is
[tex]E[X]=\displaystyle\sum_xx\,f(x)=3.16[/tex]
Since X is the number of absent students on Monday, the expectation E[X] is the number of students you can expect to be absent on average on any given Monday. According to the distribution, you can expect around 3 students to be consistently absent.
c) The variance is
[tex]V[X]=E[(X-E[X])^2]=E[X^2]-E[X]^2[/tex]
where
[tex]E[X^2]=\displaystyle\sum_xx^2\,f(x)=11.58[/tex]
So the variance is
[tex]V[X]=11.58-3.16^2\approx1.59[/tex]
The standard deviation is the square root of the variance:
[tex]\sqrt{V[X]}\approx1.26[/tex]
d) Since [tex]Y=7X+3[/tex] is a linear combination of [tex]X[/tex], computing the expectation and variance of Y is easy:
[tex]E[Y]=E[7X+3]=7E[X]+3=25.12[/tex]
[tex]V[Y]=V[7X+3]=7^2V[X]\approx78.13[/tex]
e) The covariance of X and Y is
[tex]\mathrm{Cov}[X,Y]=E[(X-E[X])(Y-E[Y])]=E[XY]-E[X]E[Y][/tex]
We have
[tex]XY=X(7X+3)=7X^2+3X[/tex]
so
[tex]E[XY]=E[7X^2+3X]=7E[X^2]+3E[X]=90.54[/tex]
Then the covariance is
[tex]\mathrm{Cov}[X,Y]=90.54-3.16\cdot25.12\approx11.16[/tex]
f) Dividing the covariance by the variance of X gives
[tex]\dfrac{\mathrm{Cov}[X,Y]}{V[X]}\approx\dfrac{11.16}{1.59}\approx0.9638[/tex]
The probability that more than 3 students are absent for the probability distribution is 0.35 and the expected value of the random variable X is 3.16.
What is probability distribution?
Probability distribution is the statistical model which represent all the achievable and similar values of a random variable that it can possess in a specified range.
The probability distribution of the number of students absent on Mondays, is as follows:
- X 0 1 2 3 4 5 6 7
- f(x) 0.02 0.03 0.26 0.34 0.22 0.08 0.04 0.01
- (a) The probability that more than 3 students are absent.
The probability that more than 3 students are absent is equal to the sum of all the prbabilites which is more than number 3. Thus,
[tex]P=f(4)+f(5)+f(6)+f(7)\\P=0.22+ 0.08 +0.04 +0.01\\ P=0.35[/tex]
- (b) The expected value of the random variable X . Interpret this expected value.
The expected value is,
[tex]\mu=\sum x.f(x)\\\mu=0.02 + 0.03 +0.26 +0.34 +0.22 +0.08 +0.04 +0.01\\\mu=3.16[/tex]
- (c) Compute the variance and standard deviation of the random variable X .
Standard daviation of the random variable X ,
[tex]\sigma=\sqrt{\sum x^2.f(x)-\mu^2}\\\sigma=\sqrt{} (0^2\times0.02 + 1^2\times0.03 +2^2\times0.26 +3^2\times0.34 +4^2\times0.22 +5^2\times0.08 +6^2\times0.04 +7^2\times0.01-3.16^2)\\\sigma=\sqrt{1.594}\\\sigma\approx1.2627[/tex]
The value of variance is,
[tex]V[X]=\sigma^2\\V[X]=1.2627^2\\V[X]=1.59[/tex]
- (d) The expected value and variance of Y=7X+3-
The expected value is,
[tex]E[Y]=E[7X+3]\\E[Y]=7E[X]+3\\E[Y]=7\times3.16+3\\E[Y]=25.12[/tex]
The variance of Y=7X+3 is,
[tex]V[Y]=V[7X+3]\\V[Y]=7^2V[X]\\V[Y]=7\times1.59\\V[Y]=77.91[/tex]
- (e) The covariance Cov(X,Y)-
[tex]Cov(X,Y)=E[7x^2+3x]-E[X]E[Y]\\ Cov(X,Y)=7E[X^2]3E[X]-E[X]E[Y]\\ Cov(X,Y)=90.54-3.16\times25.12\\Cov(X,Y)=11.16[/tex]
- (f) The value of the ratio Cov(X,Y)/Var(X) ?
[tex]\dfrac{Cov[X,Y]}{V[X]}=\dfrac{11.16}{1.59}\\\dfrac{Cov[X,Y]}{V[X]}=0.9638[/tex]
Hence, the probability that more than 3 students are absent for the probability distribution is 0.35 and the expected value of the random variable X is 3.16.
Learn more about the probability distribution here;
https://brainly.com/question/26615262
