Suppose X Is Normally Distributed With A Mean Of 20 And A Standard Deviation Of 4. What Is P X

The mean and variance of a binomial distribution are 4 and 3 respectively. What is the probability of obtaining 5 successes?

For a binomial distribution, E(X) = np and Var(X) = npq (where p is the probability of success and q is the probability of failure where q = 1 - p). Note the following:[math]np = 3[/math][math]npq = \frac{3}{2}[/math]Thus, we have that [math]3q = \frac{3}{2}[/math] and thus [math]q = \frac{1}{2}[/math]. Since we know that q = 1-p, then [math]p = \frac{1}{2}[/math]. Plugging the value for p into np = 3, we get that n = 6. Now that we know the values for n, p and q, we can plug these into the binomial mass function so that we can find the probability of 5 successes. The binomial mass function is[math]P(X = x) = \binom{n}{x}p^{x}q^{n-x}[/math]and so with n = 6, [math]p = \frac{1}{2}[/math], [math]q = \frac{1}{2}[/math] and x = 5, we have the following:[math]P(X = 5) = \binom{6}{5}(0.5)^{5}(0.5)^{1}[/math][math] = \frac{3}{32}[/math]Therefore, the probability of five successes is 0.09375.

X is a uniformly distributed random variable that takes values from 0 and 1. What is the value of [math]E{X^3}[/math]?

I am assuming a continuous uniform distribution where,[math] U ~ (a,b) \equiv U ~ (0,1) [/math]Therefore,[math] E [X^3] = \int_a^b x^3 . f(x) dx [/math][math]            = \int_0^1 x^3 . f(x) dx [/math]where, [math] f(x) = 1/(b-a) \Longrightarrow f(x) = 1/(1-0) = 1 [/math]Therefore [math]  E [X^3] = \int_0^1 x^3 . 1 . dx  [/math]                [math]           = (x^4 / 4) |(0,1) = 1/4 = 0.25 [/math]

The mean of a data is 10. If each observation is multiplied by 5 and then 1 is added to each result, how can you find the mean of the new observations obtained?

A2AMean = S/n (sum of all no.)S = a1+a2+….5*a1+ 5a2+… =5(a1+a2+..) = 5Sagain 1 is added t0 every no.hence ,5a1+1 +5a2+1 + 5a3+1 +….. n terms = 5(a1 + a2+ a3…) + 1+1+1… n terms = 5S +nnew mean = (5S+n)/n = 5* Mean + 1 = 51