Question

if a discrete random variable x with the following distribution has mean value of 2.2. find: part a: the value of m and the value of n part b: p(x < 4) part c: variance of x

Explanation:

Step1: Use the property of probability distribution sum

The sum of all probabilities in a probability - distribution is 1. So, $m + 0.1+0.3 + n+0.3=1$, which simplifies to $m + n=0.3$.

Step2: Use the formula for the mean of a discrete random variable

The mean $\mu=\sum_{i}x_ip_i$. So, $\mu=- 5m+1\times0.1 + 2\times0.3+4n + 7\times0.3$. Given $\mu = 2.2$, we have $-5m + 0.1+0.6 + 4n+2.1 = 2.2$, which simplifies to $-5m+4n=-0.6$.

Step3: Solve the system of equations for m and n

We have the system of equations

$$\begin{cases}m + n=0.3\\-5m + 4n=-0.6\end{cases}$$

. From the first equation $m = 0.3 - n$. Substitute $m = 0.3 - n$ into the second equation: $-5(0.3 - n)+4n=-0.6$. Expand to get $-1.5 + 5n+4n=-0.6$. Combine like - terms: $9n=0.9$, so $n = 0.1$. Substitute $n = 0.1$ into $m = 0.3 - n$, we get $m = 0.2$.

Step4: Calculate $P(X\lt4)$

$P(X\lt4)=P(X=-5)+P(X = 1)+P(X = 2)=m + 0.1+0.3$. Substitute $m = 0.2$, we get $P(X\lt4)=0.2 + 0.1+0.3=0.6$.

Step5: Calculate the variance of X

The variance $\sigma^{2}=\sum_{i}(x_i-\mu)^2p_i$. First, calculate $(x_i-\mu)$ for each $x_i$:
When $x=-5$, $(-5 - 2.2)^2\times0.2=( - 7.2)^2\times0.2 = 51.84\times0.2 = 10.368$.
When $x = 1$, $(1 - 2.2)^2\times0.1=(-1.2)^2\times0.1 = 0.144$.
When $x = 2$, $(2 - 2.2)^2\times0.3=(-0.2)^2\times0.3 = 0.012$.
When $x = 4$, $(4 - 2.2)^2\times0.1=(1.8)^2\times0.1 = 0.324$.
When $x = 7$, $(7 - 2.2)^2\times0.3=(4.8)^2\times0.3 = 6.912$.
$\sigma^{2}=10.368+0.144 + 0.012+0.324+6.912=17.76$.

Answer:

Part a: $m = 0.2$, $n = 0.1$
Part b: $P(X\lt4)=0.6$
Part c: $17.76$