Random Variable Probability Calculation
This problem requires finding the ratio $\frac{a}{b}$ for a random variable $X$ using its probability distribution and a condition involving its mean ($\mu$) and standard deviation ($\sigma$). The condition is $\sigma^2 + \mu^2 = 2$. The variable $X$ takes values $0, 1, 2, 3$.
Step 1: Probability Sum Constraint
The total probability must sum to 1. This provides a relationship between $a$ and $b$. $ P(X=0) + P(X=1) + P(X=2) + P(X=3) = 1 $ $ \frac{2a+1}{30} + \frac{8a-1}{30} + \frac{4a+1}{30} + b = 1 $ Combine the fractions:
- $ \frac{(2a+1) + (8a-1) + (4a+1)}{30} + b = 1 $
- $ \frac{14a+1}{30} + b = 1 $
Multiply by 30:
- $ 14a + 1 + 30b = 30 $
- $ 14a + 30b = 29 $ (Equation 1)
Step 2: Calculate Mean $E[X]$
The mean $\mu$ (or $E[X]$) is calculated as the sum of each value multiplied by its probability.
- $ \mu = E[X] = \sum_{i=0}^{3} x_i P(X=x_i) $
- $ E[X] = \left(0 \times \frac{2a+1}{30}\right) + \left(1 \times \frac{8a-1}{30}\right) + \left(2 \times \frac{4a+1}{30}\right) + \left(3 \times b\right) $
- $ E[X] = 0 + \frac{8a-1}{30} + \frac{8a+2}{30} + 3b $
- $ E[X] = \frac{16a+1}{30} + 3b $
From Equation 1, we know $30b = 29 - 14a$, so $3b = \frac{29 - 14a}{10}$. Substitute this into the expression for $E[X]$:
- $ E[X] = \frac{16a+1}{30} + \frac{3(29 - 14a)}{30} $
- $ E[X] = \frac{16a+1 + 87 - 42a}{30} = \frac{88 - 26a}{30} $
- $ \mu = E[X] = \frac{44 - 13a}{15} $
Step 3: Calculate $E[X^2]$
The second moment, $E[X^2]$, is calculated similarly using squared values.
- $ E[X^2] = \sum_{i=0}^{3} x_i^2 P(X=x_i) $
- $ E[X^2] = \left(0^2 \times \frac{2a+1}{30}\right) + \left(1^2 \times \frac{8a-1}{30}\right) + \left(2^2 \times \frac{4a+1}{30}\right) + \left(3^2 \times b\right) $
- $ E[X^2] = 0 + \frac{8a-1}{30} + 4 \times \frac{4a+1}{30} + 9b $
- $ E[X^2] = \frac{8a-1 + 16a+4}{30} + 9b = \frac{24a+3}{30} + 9b $
Substitute $b = \frac{29-14a}{30}$ from Equation 1:
- $ E[X^2] = \frac{24a+3}{30} + 9 \left(\frac{29-14a}{30}\right) $
- $ E[X^2] = \frac{24a+3 + 261 - 126a}{30} = \frac{264 - 102a}{30} $
- $ E[X^2] = \frac{44 - 17a}{5} $
Step 4: Apply Condition $\sigma^2 + \mu^2 = 2$
The variance is defined as $\sigma^2 = Var(X) = E[X^2] - \mu^2$.
Substituting this definition into the given condition $\sigma^2 + \mu^2 = 2$ simplifies the equation:
- $ (E[X^2] - \mu^2) + \mu^2 = 2 $
- $ E[X^2] = 2 $
Step 5: Solve for Parameter $a$
Equate the calculated $E[X^2]$ to 2 and solve for $a$.
- $ \frac{44 - 17a}{5} = 2 $
- $ 44 - 17a = 10 $
- $ 17a = 44 - 10 = 34 $
- $ a = \frac{34}{17} = 2 $
Step 6: Solve for Parameter $b$
Substitute the value $a=2$ into Equation 1 ($14a + 30b = 29$).
- $ 14(2) + 30b = 29 $
- $ 28 + 30b = 29 $
- $ 30b = 1 $
- $ b = \frac{1}{30} $
Step 7: Calculate the Ratio $\frac{a}{b}$
Finally, compute the ratio $\frac{a}{b}$ using the found values $a=2$ and $b = \frac{1}{30}$.
- $ \frac{a}{b} = \frac{2}{1/30} $
- $ \frac{a}{b} = 2 \times 30 = 60 $
Final Answer Calculation
The calculation shows that $\frac{a}{b} = 60$. This value corresponds to Option C.
Similar Questions
- Let $X = \{x \in \mathbb{N} : 1 \leq x \leq 19\}$ and for some $a, b \in \mathbb{R}, Y = \{ax + b : x \in X\}$. If the mean and variance of the elements of $Y$ are 30 and 750, respectively, then the sum of all possible values of $b$ is
If the mean and median of the data
x 0-10 10-20 20-30 30-40 40-50 f 3 6 2 x y $\Sigma f = 20$ are equal, then $xy^2$ is equal to
If the mean of the data: 7,8,9,7,8,7,$\lambda$,8 is 8, then the variance of this data is :-
Four dice are thrown simultaneously and the numbers shown on these dice are recorded in $2\times2$ matrices. The probability that such formed matrices have all different entries and are non-singular, is :
Consider the following frequency distribution :
Class 0 - 6 6 - 12 12 - 18 18 - 24 24 - 30 Frequency a b 12 9 5 If mean = $\frac{309}{22}$ and median=$14$, then the value $(a - b)^2$ is equal to ________.
- For 10 observations $x_1, x_2, \dots, x_{10}$, if $\sum_{i=1}^{10} (x_i + 2)^2 = 180$ and $\sum_{i=1}^{10} (x_i - 1)^2 = 90$, then their standard deviation is:
- The mean and variance of a data of 10 observations are 10 and 2, respectively. If an observation $\alpha$ in this data is replaced by $\beta$, then the mean and variance become 10.1 and 1.99, respectively. Then $\alpha + \beta$ equals
- The mean and standard deviation of 100 observations are 40 and 5.1, respectively. By mistakeone observation is taken as 50 instead of 40. If the correct mean and the correct standarddeviation are $\mu$ and $\sigma$ respectively, then $10(\mu +\sigma)$ is equal to
- The probability distribution of a random variable X is given below :
X $4\text{k}$ $\frac{30}{7}\text{k}$ $\frac{32}{7}\text{k}$ $\frac{34}{7}\text{k}$ $\frac{36}{7}\text{k}$ $\frac{38}{7}\text{k}$ $\frac{40}{7}\text{k}$ $6\text{k}$ P(X) $\frac{2}{15}$ $\frac{1}{15}$ $\frac{2}{15}$ $\frac{1}{5}$ $\frac{1}{15}$ $\frac{2}{15}$ $\frac{1}{5}$ $\frac{1}{15}$
If $\text{E}(\text{X}) = \frac{263}{15}$, then $\text{P}(\text{X} < 20)$ is equal to : Let the mean and variance of 8 numbers $-10, -7, -1, x, y, 9, 2, 16$ be $\frac{7}{2}$ and $\frac{293}{4}$, respectively.
Then the mean of 4 numbers $x, y, x + y + 1, |x - y|$ is :
Important Questions from Measures of Dispersion and Probability
- Let $X = \{x \in \mathbb{N} : 1 \leq x \leq 19\}$ and for some $a, b \in \mathbb{R}, Y = \{ax + b : x \in X\}$. If the mean and variance of the elements of $Y$ are 30 and 750, respectively, then the sum of all possible values of $b$ is
If the mean and median of the data
x 0-10 10-20 20-30 30-40 40-50 f 3 6 2 x y $\Sigma f = 20$ are equal, then $xy^2$ is equal to
If the mean of the data: 7,8,9,7,8,7,$\lambda$,8 is 8, then the variance of this data is :-
Four dice are thrown simultaneously and the numbers shown on these dice are recorded in $2\times2$ matrices. The probability that such formed matrices have all different entries and are non-singular, is :
Consider the following frequency distribution :
Class 0 - 6 6 - 12 12 - 18 18 - 24 24 - 30 Frequency a b 12 9 5 If mean = $\frac{309}{22}$ and median=$14$, then the value $(a - b)^2$ is equal to ________.
