Download App

Probability — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsProbability4 Marks
Question
The probability distribution of a discrete random variable X is given as under:
$\text{X}$
 $1$
 $2$
 $4$
 $2\text{A}$
 $3\text{A}$
 $5\text{A}$
$\text{P}(\text{X})$
 $\frac{1}{2}$
 $\frac{1}{5}$
 $\frac{3}{25}$
 $\frac{1}{10}$
 $\frac{1}{25}$
 $\frac{1}{25}$
Calculate:
  1. The value of A if E(X) = 2.94
  2. Variance of X.
✓

Answer

We have
$\sum\text{X}\text{P}(\text{X})=\frac{1}{2}+\frac{2}{5}+\frac{2}{25}+\frac{2\text{A}}{10}+\frac{3\text{A}}{25}+\frac{5\text{A}}{25}$
$=\frac{25+20+24+10\text{A}+6\text{A}+10\text{A}}{50}$
$=\frac{69+26\text{A}}{50}$
Since, $\text{E}(\text{X})=\sum\text{X}\text{P}(\text{X})$
$\Rightarrow2.94=\frac{69+26\text{A}}{50}$
$\Rightarrow26\text{A}=50\times2.94-69$
$\Rightarrow\text{A}=\frac{147-69}{26}$
$=\frac{78}{26}=3$
We know that,
$\text{Var}(\text{X})=\text{E}(\text{X}^2)-[\text{E}(\text{X})]^2$
$=\sum\text{X}^2\text{P}(\text{X})-[\text{E}(\text{X})]^2$
$=\frac{1}{2}+\frac{4}{5}+\frac{48}{25}+\frac{4\text{A}^2}{10}+\frac{9\text{A}^2}{25}+\frac{25\text{A}^2}{25}-[\text{E}(\text{X})]^2$
$=\frac{25+40+96+20\text{A}^2+18\text{A}^2+50\text{A}^2}{50}-[\text{E}(\text{X})]^2$
$=\frac{161+88\text{A}^2}{50}-[\text{E}(\text{X})]^2$
$=\frac{161+88\times(3)^2}{50}-[\text{E}(\text{X})]^2$
$=\frac{953}{50}-[2.94]^2$ $[\because\text{E}(\text{X})=2.94]$
$=19.06-8.6436$
$=10.4164$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "4 Marks" from Probability→Probability — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

Evaluate the following integrals:
$\int_{0}^\limits{1}\frac{1}{1+2\text{x}+2\text{x}^2+2\text{x}^3+\text{x}^4}\text{ dx}$
→
Solve the following differential equation:
$\text{x}\cos\Big(\frac{\text{y}}{\text{x}}\Big).(\text{y dx + x dy})=\text{y}\sin\Big(\frac{\text{y}}{\text{x}}\Big).(\text{x dy}-\text{y dx})$
→
Find the vector and cartesian forms of the plane passing through the point (1, 2, -4) and parallel to the lines $\vec{\text{r}}=(\hat{\text{i}}+2\hat{\text{j}}-4\hat{\text{k}})+\lambda(2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}})$ and $\vec{\text{r}}=(\hat{\text{i}}-3\hat{\text{j}}+5\hat{\text{k}})+\mu(\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}).$ Also, find the distance of the point (9, -8, -10) from the plane thus obtained.
→
Solve the Linear Programming Problem graphically:
Maximize Z = 5x + 3y subject to 3x + 5y $\leq$ 15, 5x + 2y $\leq$ 10, x $\geq$ 0, y $\geq$ 0.
→
Evaluvate the following intregals:
$\int\frac{1}{\text{x}(\text{x}-2)(\text{x}-4)}\ \text{dx}$
→
Without expanding, show that the values of the following determinant are zero:
$\begin{vmatrix}(2^{\text{x}}+2^{-\text{x}})^2&(2^{\text{x}}-2^{-\text{x}})^2&1\$3^{\text{x}}+3^{-\text{x}})^2&(3^{\text{x}}-3^{-\text{x}})^2&1\$4^{\text{x}}+4^{-\text{x}})^2&(4^{\text{x}}-4^{-\text{x}})^2&1\end{vmatrix}$
→
Show that the sum of three vectors determined by the medians of a triangle directed from the vertices is zero.
→
Form the differential equation of the family of circles in the first quadrant which touch the coordinate axes.
→
If the derivative of tan-1 (a + bx) takes the value 1 at x = 0, prove that 1 + a2 = b.
→
Evaluate the following definite integrals:
$\int\limits_{\frac{\pi}{3}}^{\frac{\pi}{4}}(\tan\text{x}+\cot\text{x})^2\text{ dx}$
→