JEE Main Study Notes for Statistics: Statistics is the science that deals with collecting, analyzing, and presenting any data. Statistics takes the values of some parameters and plots them in a meaningful manner. The probability of getting at least one question from statistics in JEE Main Mathematics Question Paper is 100%. Candidates should ensure that they start their preparation early. Check JEE Main Preparation Tips.
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A Central Tendency is a central value of a probability distribution. The measures indicate where most values fall in the distribution. In statistics, there are two types of averages under which the central measures of tendency fall:
If n numbers are x1, x2, x3 ........ xn then their arithmetic mean is
Let f1, f2, ........, fn is the corresponding frequencies of x1, x2, .......xn. Then
Class Mark of the class interval a – b,
For classified data, we take the class marks x1, x2, ...., xn of the classes as variables and
Must Read:
If x1, x2, ......, xn be n values of the variable then
For Frequency Distribution , where
or
For Frequency Distribution, , where
Where, a = assumed mean, d = x – a, n = no. of terms
Where a = assumed mean, d = x – a, f = frequency of variable x
Where a = assumed mean, d = x – a, x = class-mark of the class-interval, f = frequency of the class interval
Also Read:
Median is the middle-most value of the data given when arranged in ascending or descending order. If the question given is in the form of ungrouped data, we first arrange the data into ascending or descending order and then identify the Median by applying the formulas in accordance with the data.
Arrange the data provided in ascending or descending order and then find the n (number of terms).
and term. Hence, Median = Mean of and terms.
At times the questions are asked in the form of Overlapping intervals, e.g. 10 – 20, 20 – 30, 30 – 40, where the Upper limit for 10 - 20 interval = 20, and Lower limit = 10
For Non-overlapping intervals, e.g. 10 - 19, 20 - 29, Upper boundary for 10 - 19 = = lower boundary of 20 – 29.
To determine the median with the help of graphs, draw the “less than” ogive and “more than” ogive for the distribution. The abscissa of the point of intersection of these ogives is the median.
Mode is the variable that occurs the maximum number of times in a given series of elements. It is the value at the point in the distribution where items tend to be most heavily concentrated.
Mode from the following numbers of a variable 70, 80, 90, 96, 70, 96, 96, 90 is 96 as 96 occurs a maximum number of times.
In the above-given data, the frequency that has appeared a maximum number of times is 5, with 13 occurrences. Hence, the model is 5.
The class that has maximum frequency is called the MODAL CLASS and the middle point of the modal class is called the CRUDE MODE. The class just before the modal class is called PRE-MODAL CLASS and the class after the modal class is called the POST-MODAL CLASS
Determination of mode for classified Data (continuous distribution)
Must Read: JEE Main Study Notes on Differential Calculus
Dispersion normally means the extent to which the numerical data tends to spread. It is also called Variation. Its measurement is called Deviation. There are four Measures (methods) of Dispersion:
Also called Mean Absolute Deviation, it is the mean of the absolute deviations of a set of data from its mean denoted by or or
where n = no. of terms, z = A or M or M0
where x stands for class-mark
It measures the dispersion of the dataset relative to its mean. It is calculated as the square root of the variance. (The square of S.D., i.e., σ2 is called the Variance.)
where x = class-mark of the interval.
Read: JEE Main Probability Study Notes
Question: Compute the median from the following table.
Marks obtained | No. of students |
---|---|
0-10 | 2 |
10-20 | 18 |
20-30 | 30 |
30-40 | 45 |
40-50 | 35 |
50-60 | 20 |
60-70 | 6 |
70-80 | 3 |
Answer:
Marks obtained | No. of students | Cumulative frequency |
---|---|---|
0-10 | 2 | 2 |
10-20 | 18 | 20 |
20-30 | 30 | 50 |
30-40 | 45 | 95 |
40-50 | 35 | 130 |
50-60 | 20 | 150 |
60-70 | 6 | 156 |
70-80 | 3 | 159 |
N = Σf = 159 (Odd number)
Median is [1 / 2] (n + 1) = [1 / 2] [(159 + 1)] = 80th value, which lies in the class [30 – 40] (see the row of cumulative frequency 95, which contains 80).
Hence the median class is [30 – 40].
We have l = Lower limit of median class=30
f = frequency of median class=45
C = Total of all frequencies preceding median class=50
i = width of class interval of median class=10
Required median = l + ([N / 2 − C] / f) * i
= 30 + ([159 / 2 − 50] / 45) × 10
= 3 + 295 / 45
= 36.55
Question: The mean and S.D. of the marks of 200 candidates were found to be 40 and 15 respectively. Later, it was discovered that a score of 40 was wrongly read as 50. The correct mean and S.D. respectively are ______________.
Answer: Corrected Σx = 40 × 200 − 50 + 40 = 7990
Corrected x bar = 7990 / 200 = 39.95
Incorrect Σx2 = n [σ2 + (\bar{x}^2xˉ2) = 200 [152 + 402] = 365000
Correct Σx2 = 365000 − 2500 + 1600 = 364100
Corrected σ = \sqrt{\frac{364100 }{ 200}}200364100 − (39.95)2
= √(1820.5 − 1596)
= √224.5
= 14.98
Question: The average of n numbers x1, x2,……. xn is M. If xn is replaced by x′, then what is the new average?
Answer: M = \frac{x_1, x_2,……. x_n}{n}nx1,x2,…….xn i.e.,
nM = x1, x2,……. xn−1 + xn
nM – xn = x1, x2,……. xn−1
[nM – xn + x′] / n = [x1, x2,……. xn−1 + x′] / n
New average = [nM − xn + x′] / n
Question: The following data gives the distribution of the height of students.
Height (in cm) | 160 | 150 | 152 | 152 | 161 | 154 | 155 |
Name of students | 12 | 8 | 4 | 4 | 3 | 3 | 7 |
Question: What is the median of the distribution?
Answer: Arranging the data in ascending order of magnitude, we obtain
Height (in cm) | 150 | 152 | 154 | 155 | 156 | 160 | 161 |
Number of students | 8 | 4 | 3 | 7 | 3 | 12 | 4 |
Cumulative frequency | 8 | 12 | 15 | 22 | 25 | 37 | 41 |
Here, the total number of items is 41 i.e., an odd number.
Hence, the median is [(41 + 1) / 2]th i.e., 21st item.
From the cumulative frequency table, we find that the median i.eThe 21st item is 155.
(All items from 16 to 22nd are equal, each 155).
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