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Chapter 18: Problem 697

If \(x\) and \(y\) are related as \(2 x+5 y=15\) and mean deviation of y about mean is 10 then the mean deviation of \(x\) about mean is (a) 25 (b) 50 (c) 20 (d) 25

Short Answer

Expert verified
The mean deviation of x about its mean is 25, which is given by option (a).

Step by step solution

01

Express x in terms of y

From the given equation, we can find the relationship between x and y: \(2x + 5y = 15\) Now, we need to express x in terms of y: \(2x = 15 - 5y\) Divide both sides by 2: \(x = \frac{15 - 5y}{2}\)
02

Compute the mean deviation of x

We are given that the mean deviation of y about its mean is 10. Recall that mean deviation is the average of absolute differences between the values and their mean. Since the relationship between x and y is linear, their mean deviations are proportional to each other, and we have: Mean Deviation of x = \(\left|\frac{d(\frac{15-5y}{2})}{dy}\right|\) × Mean Deviation of y Take the derivative of x with respect to y: \( \frac{d(\frac{15-5y}{2})}{dy} = -\frac{5}{2}\) The absolute value of the derivative is: \(\left|-\frac{5}{2}\right| = \frac{5}{2}\) Now multiply by the given mean deviation of y: Mean Deviation of x = \(\frac{5}{2}\) × 10 Mean Deviation of x = 25
03

Find the answer in the given choices

The mean deviation of x about its mean is 25. Looking at the given choices, we can see that the correct answer is: (a) 25

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Most popular questions from this chapter

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) \(\sqrt{10}\)

A \(2 \times 2\) determinant is such that all its entries are \(1,-1\) or \(0 .\) If one determinant is chosen from such determinants what is the probability that the value of the determinant is zero? (a) \((3 / 8)\) (b) \((11 / 27)\) (c) \((2 / 9)\) (d) \((25 / 81)\)

Two numbers from \(S=\\{1,2,3,4,5,6\\}\) are selected one by one without replacement. The probability that minimum of the two numbers is less than 4 is (a) \((1 / 15)\) (b) \((14 / 15)\) (c) \((1 / 5)\) (d) \((4 / 5)\)

The A.M. of 9 terms is 15 . If one more term is added to this series then the A.M. becomes 16 . The value of added term is (a) 30 (b) 27 (c) 25 (d) 23

Find average deviation from median for given frequency distributions $$ \begin{array}{|l|c|c|c|c|c|c|} \hline \text { Class } & 0-10 & 10-20 & 20-30 & 30-40 & 40-50 & 50-60 \\ \hline \text { Frequency } \mathrm{f} & 6 & 7 & 15 & 16 & 4 & 2 \\ \hline \end{array} $$ (a) \(10.16\) (b) \(16.10\) (c) \(10.10\) (d) \(16.16\)
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