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Chapter 18: Problem 686

If \(x\) and \(y\) are related as \(4 x-3 y=10\) and the mean deviation of \(\mathrm{x}\) is 10 then the mean deviation of \(\mathrm{y}\) is (a) 13 (b) \(12.3\) (c) \(13.3\) (d) \(13.5\)

Short Answer

Expert verified
The mean deviation of y is 13.3.

Step by step solution

01

Find the relationship between the mean deviations of x and y

Since \(4x - 3y = 10\), we can say that \(y = \frac{4}{3}x - \frac{10}{3}\). Let the mean deviation of x be \(M_x\) and the mean deviation of y be \(M_y\). Let \(x_1\) and \(x_2\) be any two values of x such that their mean deviation is 10. Then, \(M_x = |x_1 - x_2| = 10\) Now, we need to find the corresponding values of y for these x values, i.e., \(y_1\) and \(y_2\). We can do that using the relationship we found. \(y_1 = \frac{4}{3}x_1 - \frac{10}{3}\) \(y_2 = \frac{4}{3}x_2 - \frac{10}{3}\)
02

Calculate the mean deviation of y using the relationship

Now, let's find the mean deviation of y using the values of \(y_1\) and \(y_2\): \(M_y = |y_1 - y_2|\) Now substitute the values of \(y_1\) and \(y_2\) we found in step 1: \(M_y = \left|\left(\frac{4}{3}x_1 - \frac{10}{3}\right) - \left(\frac{4}{3}x_2 - \frac{10}{3}\right)\right|\) Simplify the expression: \(M_y = \left|\frac{4}{3}(x_1 - x_2)\right|\) Since we know \(|x_1 - x_2| = M_x = 10\), we can substitute it in the expression: \(M_y = \frac{4}{3}(10)\)
03

Find the mean deviation of y

Multiply the fraction and the constant to find the mean deviation of y: \(M_y = 4\left(\frac{10}{3}\right) = \frac{40}{3} = 13.3\) So, the mean deviation of y is \(13.3\). Therefore, the correct answer is (c) 13.3.

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Most popular questions from this chapter

Variance of \(1,3,5,7 \ldots \ldots \ldots(4 n+1)\) is (a) \([\\{2 n(2 n-1)\\} / 3]\) (c) \((1 / n) \sqrt\left[\left(n^{2}-1\right) / 3\right] 100\) (d) \([\\{4 n(n+1)\\} / 3]\)

If \(n\) integers taken at random are multiplied together, then the probability that the last digit of the product is \(1,3,7\) or 9 is (a) \(\left(2^{n} / 5^{n}\right)\) (b) \(\left[\left(4^{n}-2^{n}\right) / 5^{n}\right]\) (c) \(\left(4^{n} / 5^{n}\right)\) (d) \((2 / 5)\)

If \(\quad 10 \Sigma_{i=1}(x i-8)=9\) and \({ }^{10} \sum_{i=1}(x i-8)^{2}=45\) then standard deviation of \(\mathrm{x}_{1}, \mathrm{x}_{2}, \mathrm{x}_{3} \ldots \ldots \ldots \mathrm{x}_{10}\) is (a) \(19.2\) (b) \(12.92\) (c) \(1.82\) (d) \(1.92\)

If the variance of \(\mathrm{x}\) is 4 then the variance of \(3+5 \mathrm{x}\) is (a) 100 (b) 103 (c) 20 (d) 23

For observations \(x_{1}, x_{2}, \ldots \ldots . . x_{n}, n_{i=1}(x i+4)=100\) and \(n_{i=1}(x i+6)=140\) then \(n=\) and \(\underline{x}=\) (a) 3,20 (b) 20,3 (c) 1,20 (d) 20,1
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