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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

The sum of the squares of deviation for 10 observations taken from their mean 30 is 90 . The coefficient of variation is (a) \(20 \%\) (b) \(10 \%\) (c) \(11 \%\) (d) \(12 \%\)

There are two boxes. Box I contains 4 Red and 3 white balls. Box II contains 5 red and 2 white balls. Two balls are transferred from Box I to Box II. One ball is then drawn from box II randomly. What is the probability for that ball to be red? (a) \((43 / 63)\) (b) \((23 / 73)\) (c) \((34 / 63)\) (d) None

For observations \(x_{1}, x_{2}, \ldots \ldots . . x_{n}\). If \(\sum_{i=1}(x i+1)^{2}=9 n\) and \({ }^{n} \sum_{i=1}(x i-1)^{2}=5 n\) then standard deviation of the data is (a) \(\sqrt{3}\) (b) \(\sqrt{5}\) (d) \(\sqrt{2}\) (d) \(\sqrt{10}\)

12 balls are distributed among three boxes. The probability that the first box contain 3 balls is (a) \((110 / 9)(2 / 3)^{10}\) (b) \((9 / 110)(2 / 3)^{10}\) (c) \(\left[\left({ }^{12} 3\right) / 12^{3}\right] \cdot 2^{9}\) (d) \(\left[\left(\begin{array}{ll}12 & 3\end{array}\right) / 3^{12}\right]\)

The mean of the series \(a, a+d, a+2 d \ldots \ldots . a+(2 n+1) d\) is (a) \(a+[(2 n+1) / 2] d\) (b) \(a+(n+1) d\) (c) \(a+(2 n+1) d\) (d) \(a+[(2 n-1) / 2] d\)
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