2. A random variable X has a probability mass function given by: 1 0.2 a. Determine P(x) = 2 b.Find P(X<2) c. Find P(X≥ 3) X P(x) 0 0.1 2 ? 3 0.4
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- (Sec. 3.2) A student is required to enroll in one, two, three, four, five, six on the desired courseload) at a local university. Let Y the number of classes the next student enrolls themselves in. The probability that y classes are selected is known to be proportional to y+1, in other words the pmf of Y is given by p(y) = k(y+1) for y 1,...,7, and 0 otherwise (a) What is the value of k? or seven classes (depending (b) What is the probability that at most four classes are enrolled in? (c) What is the probability that a student enrolls in between three and five classes (inclusive)? y? /40 for y 1,.,7 be the pmf of Y? Explain why why not (d) Could p(y) orAn elevator has a placard stating that the maximum capacity is 1630 lb—10 passengers. So,10 adult male passengers can have a mean weight of up to 1630/10=163 pounds.If the elevator is loaded with 10 adult male passengers, find the probability that it is overloaded because they have a mean weight greater than 163 lb. (Assume that weights of males are normally distributed with a mean of 170 lb and a standard deviation of 29 lb.) Does this elevator appear to be safe? The probability the elevator is overloaded is.A- Let x be a discrete random variable with probability distribution function f(x)=k( x2 +20) and x= −1,1,2,3. Find the value of k. Find the Variance of X. B- Let x denote a discrete random variable which can take the values −2,0, and 5. Given that the expectation of X is 8/100 and P(X=−2)=8/20 , find P(X=5).
- Q3: Suppose the joint probability function of X and Y is given by x= 5, y = 0 7 x= 5, y = 3 7 x = 5, y = 4 P (x,y)- *= 8, y = 0 7 x= 8, y = 4 else Compute E(X|Y-0), E(X|Y 4), E(X|Y=3)3. Determine the value c so that each of the following functions can serve as probability distributions of the discrete random variable X: a. f(x) = c(x² +4) for x = 0,1, 2, 3; 3 b. f(x)%3Dс for x= 0,1, 2 3-x4. Find the mean of the discrete random variable X with the following probability distribution. Use this formula. μ -ΣΙΧ . P (X)] P(x) х. Р(х) x2 x² . P(x) X 14 1 /½ 5. Referring to the table above, solve for the variance and standard deviation. Use this formula. o² = E[x2 . P(X)1 – µ? E[x² - P(X)] – µ² | O =
- The possible values of a discrete random variable X are 0, 1, 3, and 6 with respective probabilities 0.2, 0.3, 0.1, 0.4. Find E[X] and Var(X).Suppose that the distribution function of X is given by: (0, b<0 b Osb<1 1 b F(b) =- D=1, Isb<2 2 11 2sb<3 12 b23 1. Determine its corresponding probability mass function. 3 2. Find1. Verify whether the following functions can be considered as probability mass functions: (i) P(x = x) = x² + 1 18 (iii) P(X= x) = -, x = 0, 1, 2, 3 (ii) P(X - x) - ²-2, -, x= 1, 2, 3 2x + 1 18 -, x = 0, 1, 2, 3 [Ans.: Yes) [Ans.: No] [Ans.: No]