Let $x _1, x _2, \ldots \ldots x _{10}$ be ten observations such that $\sum_{i=1}^{10}\left(x_i-2\right)=30, \sum_{i=1}^{10}\left(x_i-\beta\right)^2=98, \beta>2$ and their variance is $\frac{4}{5}$. If $\mu$ and $\sigma^2$ are respectively the mean and the variance of $2\left( x _1-1\right)+4 \beta, 2\left( x _2-1\right)+$ $4 \beta, \ldots . ., 2\left(x_{10}-1\right)+4 \beta$, then $\frac{\beta \mu}{\sigma^2}$ is equal to :
- [JEE MAIN 2025]
- A$100$
- B$110$
- C$120$
- D$90$
Similar Questions
Find the mean, variance and standard deviation using short-cut method
Height in cms
$70-75$
$75-80$
$80-85$
$85-90$
$90-95$
$95-100$
$100-105$
$105-110$
$110-115$
No. of children
$3$
$4$
$7$
$7$
$15$
$9$
$6$
$6$
$3$
Find the mean, variance and standard deviation using short-cut method
| Height in cms | $70-75$ | $75-80$ | $80-85$ | $85-90$ | $90-95$ | $95-100$ | $100-105$ | $105-110$ | $110-115$ |
| No. of children | $3$ | $4$ | $7$ | $7$ | $15$ | $9$ | $6$ | $6$ | $3$ |
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Let the mean and the variance of $5$ observations $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ be $\frac{24}{5}$ and $\frac{194}{25}$ respectively. If the mean and variance of the first $4$ observation are $\frac{7}{2}$ and $a$ respectively, then $\left(4 a+x_{5}\right)$ is equal to
- [JEE MAIN 2022]
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- [JEE MAIN 2022]
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