Q.9. A data consists of 20 observations . If and , then the ratio of mean to standard deviation of

Q: What is the correct answer for JEE Main Mathematics Q9?

The correct answer is B. Sol. Given $N=20$. (1) $\sum_{i=1}^{20}(x_i + 5)^2 = 2500 \Rightarrow \sum (x_i^2 + 10x_i + 25) = 2500 \Rightarrow \sum x_i^2 + 10 \sum x_i + 20 \times 25 = 2500 \Rightarrow \sum x_i^2 + 10 \sum x_i + 500 = 2500 \Rightarrow \sum x_i^2 + 10 \sum x_i = 2000$ (Eq. A) (2) $\sum_{i=1}^{20}(x_i - 5)^2 = 100 \Rightarrow \sum (x_i^2 - 10x_i + 25) = 100 \Rightarrow \sum x_i^2 - 10 \sum x_i + 20 \times 25 =

Q: Q.9. A data consists of 20 observations . If and , then the ratio of mean to standard deviation of this data is:

The correct answer is B. Sol. Given $N=20$. (1) $\sum_{i=1}^{20}(x_i + 5)^2 = 2500 \Rightarrow \sum (x_i^2 + 10x_i + 25) = 2500 \Rightarrow \sum x_i^2 + 10 \sum x_i + 20 \times 25 = 2500 \Rightarrow \sum x_i^2 + 10 \sum x_i + 500 = 2500 \Rightarrow \sum x_i^2 + 10 \sum x_i = 2000$ (Eq. A) (2) $\sum_{i=1}^{20}(x_i - 5)^2 = 100 \Rightarrow \sum (x_i^2 - 10x_i + 25) = 100 \Rightarrow \sum x_i^2 - 10 \sum x_i + 20 \times 25 =

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Q.9. A data consists of 20 observations . If  and , then the ratio of mean to standard deviation of this data is:

FirstInTest · Accepted Answer

Sol. Given $N=20$. (1) $\sum_{i=1}^{20}(x_i + 5)^2 = 2500 \Rightarrow \sum (x_i^2 + 10x_i + 25) = 2500 \Rightarrow \sum x_i^2 + 10 \sum x_i + 20 	imes 25 = 2500 \Rightarrow \sum x_i^2 + 10 \sum x_i + 500 = 2500 \Rightarrow \sum x_i^2 + 10 \sum x_i = 2000$ (Eq. A) (2) $\sum_{i=1}^{20}(x_i - 5)^2 = 100 \Rightarrow \sum (x_i^2 - 10x_i + 25) = 100 \Rightarrow \sum x_i^2 - 10 \sum x_i + 20 	imes 25 = 100 \Rightarrow \sum x_i^2 - 10 \sum x_i + 500 = 100 \Rightarrow \sum x_i^2 - 10 \sum x_i = -400$ (Eq. B) Add (Eq. A) and (Eq. B): $(\sum x_i^2 + 10 \sum x_i) + (\sum x_i^2 - 10 \sum x_i) = 2000 + (-400) \Rightarrow 2 \sum x_i^2 = 1600 \Rightarrow \sum x_i^2 = 800$. Subtract (Eq. B) from (Eq. A): $(\sum x_i^2 + 10 \sum x_i) - (\sum x_i^2 - 10 \sum x_i) = 2000 - (-400) \Rightarrow 20 \sum x_i = 2400 \Rightarrow \sum x_i = 120$. Mean $\bar{x} = \frac{\sum x_i}{N} = \frac{120}{20} = 6$. Variance $\sigma^2 = \frac{\sum x_i^2}{N} - (\bar{x})^2 = \frac{800}{20} - (6)^2 = 40 - 36 = 4$. Standard deviation $\sigma = \sqrt{4} = 2$. The ratio of mean to standard deviation is $\frac{\bar{x}}{\sigma} = \frac{6}{2} = 3:1$.

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