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Chapter 18: Problem 747

For a data there are 3n observations in which first \(n\) observations are \(a-d\), second n observation are a and last n observations are \(a+d\) and there variance is \((4 / 3)\) then \(|\mathrm{d}|=\) (a) 1 (b) \(\sqrt{2}\) (c) \(\sqrt{(2 / 3)}\) (d) \(\sqrt{(3 / 2)}\)

Short Answer

Expert verified
\( |d| = \sqrt{\frac{4}{3}} \) Solution: (d) \(\sqrt{(3 / 2)}\).

Step by step solution

01

In order to work with variance, we need to know its formula. The formula for variance is defined as: \[ \text{Variance} = \frac{\sum_{i=1}^n (x_i - \mu)^2}{n} \] where n is the number of observations, \(x_i\) is each observation, and \(\mu\) is the mean of the set. #Step 2: Calculate The Mean#

We need the mean of the dataset to use in the variance formula. The dataset contains 3n observations: n as \(a-d\), n as \(a\), and n as \(a+d\). So, we can write the mean as follows: \[ \mu = \frac{\sum_{i=1}^n (a-d) + \sum_{i=1}^n a + \sum_{i=1}^n (a+d)}{3n} \] Calculating the sums and simplifying the expression, we get: \[ \mu = \frac{3an}{3n} = a \] #Step 3: Apply The Variance Formula#
02

Now that we have the mean value, we can apply the variance formula on the dataset to get an expression for the variance: \[ \text{Variance} = \frac{\sum_{i=1}^n [(a-d)-a]^2 + \sum_{i=1}^n [a-a]^2 + \sum_{i=1}^n [(a+d)-a]^2}{3n} \] Substituting the given value of variance, \((4/3)\), and simplifying the expression, we get: \[ \frac{4}{3} = \frac{\sum_{i=1}^n d^2 + 0 + \sum_{i=1}^n d^2}{3n} \] #Step 4: Solve For d#

Next, we solve for d using the expression from Step 3: \[ \frac{4}{3} = \frac{2nd^2}{3n} \] Canceling out terms and simplifying, we get: \[ d^2 = \frac{4}{3} \] Now, we take the square root on both sides to find the absolute value of d: \[ |d| = \sqrt{\frac{4}{3}} \] Comparing the result with the given options, we find that our answer matches the option (d). Solution: (d) \(\sqrt{(3 / 2)}\).

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Most popular questions from this chapter

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

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) \(12 \%\)

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

If mean of \(\log x, \log 2 x, \log 8, \log 4 x, \log 4, \log x\) is \(\log 4\) then \(x=\) (where \(x>0)\) (a) 4 (b) 2 (c) 8 (d) 16

Suppose a population \(A\) has 50 observations \(101,102, \ldots \ldots . .150\) and another population \(\mathrm{B}\) has 50 observations \(201,202, \ldots \ldots \ldots 250\). If \(\mathrm{V}_{\mathrm{A}}\) and \(\mathrm{V}_{\mathrm{B}}\) represent the variance of the two populations respectively then \(\left(\mathrm{V}_{\mathrm{A}} / \mathrm{V}_{\mathrm{B}}\right)\) is (a) 1 (b) \((2 / 3)\) (c) \((3 / 2)\) (d) \((9 / 4)\)
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