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Chapter 14: Problem 39

If the sum of the deviations of 50 observations from 30 is 50 , then the mean of these observations is : [Jan. 12, 2019 (I)] (a) 30 (b) 51 (c) 50 (d) 31

Short Answer

Expert verified
The mean of the observations is 31.

Step by step solution

01

Understanding Deviations

The deviation of each observation from a number is the difference between the observation and that number. Here, the sum of deviations from 30 is given as 50, which means the sum of (observation - 30) for all 50 observations is 50.
02

Setting Up the Equation

Let's denote the sum of all observations as \( S \). The sum of the deviations from 30 can be expressed as \( S - 50 imes 30 \). We know this sum is equal to 50.
03

Deriving the Sum of Observations

Since the sum of deviations is 50, we set up the equation: \( S - 1500 = 50 \). Solving for \( S \), we get \( S = 1550 \).
04

Calculating the Mean

The mean of the observations is the total sum \( S \) divided by the number of observations, which is 50. Therefore, the mean is \( \frac{1550}{50} \).
05

Final Calculation

Perform the division: \( \frac{1550}{50} = 31 \). Thus, the mean of these observations is 31.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Deviations
In statistics, a deviation represents how far a single observation is from a specific reference point. Basically, it's the difference between each observation and a given number, sometimes referred to as a central point or a mean. In our exercise, the reference point is 30.
  • This means for each observation, you subtract 30 to find its deviation.
  • The problem states that the sum of all these deviations for 50 observations is equal to 50.
This helps us understand how the observations tend to cluster around the number 30. If the sum of deviations is a small number like 50 over several observations, it indicates that they are not too spread out from 30.
Sum of Observations
The sum of observations, often denoted as \( S \), is the total when all individual observations are added together. This value is crucial because it helps us calculate other important statistics, like the mean.
  • Initially, we do not have the direct sum of observations.
  • However, by knowing the sum of deviations and using a bit of algebra, we can find \( S \).
In our setup, we created a simple equation: \( S - 50 \times 30 = 50 \). Solving this equation allows us to deduce that the sum of all the observations is 1550.
Mathematical Expression
Mathematics allows us to express problems in a concise form using algebra. In this exercise, we translated the given information into a mathematical equation so we could find unknowns.
  • The equation derived was: \( S - 1500 = 50 \).
  • It's derived from the expression for sum of deviations: \( S - 50 \times 30 \).
This simplification lets us work with numbers more easily. By solving the equation for \( S \), it gives us a pathway to finding the mean of the observations. It's a beautiful example of how algebra can succinctly tie together information and reveal new insights.
Arithmetic Mean
The arithmetic mean, commonly just called the mean, is one of the simplest statistical tools to measure the center of a data set. It鈥檚 calculated by dividing the total sum of all observations by the number of observations.
  • In our case, with a total sum of \( S = 1550 \) and 50 observations, the mean is \( \frac{1550}{50} \).
  • This calculation results in a mean of 31.
The mean gives us a good middle point of the data. In practical terms, it tells us that the average observation in this dataset is 31. This can be handy in making predictions or understanding the general trend of the data.

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Most popular questions from this chapter

5 students of a class have an average height \(150 \mathrm{~cm}\) and variance \(18 \mathrm{~cm}^{2}\). A new student, whose height is \(156 \mathrm{~cm}\), joined them. The variance (in \(\mathrm{cm}^{2}\) ) of the height of these six students is: [Jan. 9, 2019 (I)] (a) 16 (b) 22 (c) 20 (d) 18

Let the sum of the first three terms of an A. P, be 39 and the sum of its last four terms be 178 . If the first term of this A.P. is 10 , then the median of the A.P. is : [Online April 10, 2015] (a) 28 (b) \(26.5\) (c) \(29.5\) (d) 31

In a series of \(2 n\) observations, half of them equal \(a\) and remaining half equal \(-a\). If the standard deviation of the observations is 2, then \(|a|\) equals. (a) \(\frac{\sqrt{2}}{n}\) (b) \(\sqrt{2}\) (c) 2 (d) \(\frac{1}{n}\)

If the mean deviation of the numbers \(1,1+\mathrm{d}, \ldots, 1+100 \mathrm{~d}\) from their mean is 255 , then a value of \(\mathrm{d}\) is : [Online April 9, 2016] (a) \(10.1\) (b) \(5.05\) (c) \(20.2\) (d) 10

The sum of 100 observations and the sum of their squares are 400 and 2475, respectively. Later on, three observations, 3,4 and 5, were found to be incorrect. If the incorrect observations are omitted, then the variance of the remaining observations is : \(\quad\) [Online April 9, 2017] (a) \(8.25\) (b) \(8.50\) (c) \(8.00\) (d) \(9.00\)
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