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Question

If the mean of 20 observations, namely $x_1, x_2, x_3, \dots, x_{20}$ is 1.414, then what is the value of $\sum_{i=1}^{20} 100(2x_i + 4)$ ?

The correct answer is

13656 

Summation Calculation from Mean

This solution details the calculation for a summation expression, using the provided mean of observations.

Step 1: Calculate Sum of Observations

We are given 20 observations ($x_1, \dots, x_{20}$) with a mean $\bar{x}=1.414$. The mean formula is $ \bar{x} = \frac{\sum_{i=1}^{n} x_i}{n} $.

Using the given values ($n=20$):

$ \sum_{i=1}^{20} x_i = n \times \bar{x} = 20 \times 1.414 = 28.28 $

Step 2: Evaluate the Summation Expression

We need to find the value of $\sum_{i=1}^{20} 100(2x_i + 4)$. We apply summation properties:

1. Factor out the constant 100:

$ 100 \sum_{i=1}^{20} (2x_i + 4) $

2. Split the summation and simplify:

$ 100 \left( \sum_{i=1}^{20} 2x_i + \sum_{i=1}^{20} 4 \right) $

3. Use the sum of $x_i$ and the sum of a constant ($ \sum_{i=1}^{n} c = n \times c $):

$ 100 \left( 2 \sum_{i=1}^{20} x_i + (20 \times 4) \right) $

4. Substitute the calculated sum $\sum_{i=1}^{20} x_i = 28.28$ and calculate:

$ 100 \left( 2 \times 28.28 + 80 \right) $ $ = 100 \left( 56.56 + 80 \right) $ $ = 100 (136.56) $ $ = 13656 $

Final Answer

The calculated value of the summation is 13656.

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Important Questions from Statistics

  1. The mean and variance of five observations are 14 and 13.2 respectively. Three of the five observations are 11, 16 and 20. What are the other two observations ?

  2. A die is thrown 10 times and obtained the following outputs :

    1, 2, 1, 1, 2, 1, 4, 6, 5, 4

     What will be the mode of data so obtained ?  

  3. Consider the following frequency distribution :

    x1235
    f4697

    What is the value of median of the distribution ?  

  4. For data -1, 1, 4, 3, 8, 12, 17, 19, 9, 11; if M is the median of first 5 observations and N is the median of last five observations, then what is the value of 4M - N ?

  5. Let P, Q, R represent mean, median and mode. If for some distribution \(5 P=4 Q=\frac{R}{2}\) then what is \(\frac{P+Q}{2 P+0.7 R}\) equal to ?

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