Exams Prep Master
Statistics refers to the branch of mathematics that concerns the collection, organization, analysis, interpretation and presentation of data. Statistics eases large amount of numerical data. In class 10, statistics consists of measures of central tendency in both grouped and ungrouped data. The chapter also includes cumulative frequency, frequency polygon, bar graphs and mainly focuses on different ways of calculating mean, median, and mode.
Also read: Definite Integral Formula
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Important Questions
Short Answer Questions
Question. In a continuous frequency distribution, if each value is increased by five, what will be the new median. Given the old median is 21. (2015)
Solution. New median = 21 + 5 = 26
Question. What will be the mean of the first ten natural numbers?
Solution. \(\hat{a} \) € - The first 10 natural numbers are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.
Mean = (1 +2 +3 +4 +5+ 6+ 7+ 8+ 9+10) / 10 = 55/10 =5.5
Question. Find the Frequency of the 30-40 class in the following distribution. (2013)
| Marks obtained | 0 or more | 10 or more | 20 or more | 30 or more | 40 or more | 50 or more |
| Number of students | 63 | 58 | 55 | 51 | 48 | 41 |
Solution. So, the Frequency of class 30 – 40 = 3
Question. \(\hat{a} \) € -Find the value of y from the following data, which is already arranged in ascending order. The given Median is 63. 20, 24, 42, y, y +2, 73, 75, 80, 99
Solution. As the number of observations is odd, so the median will be the middle term which is y+2, in the given case. So,
y + 2 = 63
y = 63 -2 = 61
Question. The weekly household expenditure of families living in a housing society is shown below. Find the upper limit for the modal class. (2014)
| Weekly expenditure | Up to 3000 | 3000-6000 | 6000-9000 | 9000-12000 | 12000-15000 |
| No. of families | 4 | 25 | 31 | 48 | 10 |
Solution. Highest frequency = 48
So the modal class=9,000 – 12,000
The upper limit of the class=12,000
Question. Find the mode in the data given below. 1,1,7,9,5,4,5,9,5,6,7,2,3,2,4
Solution. Arranging the data in ascending order – 1,1,2,2,3,4,4,5,5,5,6,7,7,9,9
As five repeats the most times in the data. Therefore 5 is the mode.
Question. The mean of the frequency distribution given below is 18.75. Find the value of P: (2012, 2017D)
| Class mark | 10 | 5 | P | 25 | 30 |
| frequency | 5 | 10 | 7 | 8 | 2 |
Solution.
| x | f | fx |
| 10 | 5 | 50 |
| 15 | 10 | 150 |
| P | 7 | 7P |
| 25 | 8 | 200 |
| 30 | 2 | 60 |
| Fi=32 | Fixi=460+7p |
=Mean =∑xifi /∑fi\(\hat{a} \) € -
=18.75=360+7P/32
600=360+7P
600-360=7P
P=240/7=34.29
Question. Find the median using the empirical formula when it is given that the mean = 30.5 and mode = 35.3. (2014)
Solution. Mode = 3(Median) – 2(Mean)
35.3 = 3(median) – 2(30.5)
35.3 = 3(median) – 61
96.3 = 3 median
Median =\(\hat{a} \) € -96.3/3\(\hat{a} \) € -= 32.1
Question. Find the mode of the following frequency distribution. (2013)
| Class | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
| Frequency | 8 | 12 | 10 | 11 | 9 |
Solution.
| Class | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
| Frequency | 8 f0 | 12 f1 | 10 f2 | 11 | 9 |
Maximum frequency = 12
Mode= l+ \(\frac{(f1?f0)}{(2f1?f0?f2)} × h \)
= \(10 + \frac{12- 8 * 10}{24-8-10}\)
=10+6.6=16.6
Long Answer Questions
Question. The mean of the given distribution is 53. the frequencies f1\(\hat{a} \) € -and f2\(\hat{a} \) € -in the classes 20-40 and 60-80 are missing. Find the missing frequencies: (2013)
| Classes | 0-20 | 20-40 | 40-60 | 60-80 | 80-100 | Total |
| Frequency | 15 | F1 | 21 | F2 | 17 | 100 |
Solution. 53+f1+f2=100
=f1+f2=100-53=47
f2=47-f1- (1)
Mean =∑xifi /∑fi\(\hat{a} \) € -=53
2370+30f1+70f2/100=53
=2730 + 70f2 + 30f1\(\hat{a} \) € -= 5300
= 30f1+ 70f2\(\hat{a} \) € -= 5300 – 2730 = 2570
=3f1\(\hat{a} \) € -+ 7f2\(\hat{a} \) € -= 257 after dividing by 10
=3f1\(\hat{a} \) € -+7(47 – f1) = 257. From 1
= 3f1\(\hat{a} \) € -– 7f1\(\hat{a} \) € -+ 329 = 257
= -4f1\(\hat{a} \) € -= -72
= f1=\(\hat{a} \) € -−72/−4\(\hat{a} \) € -= 18
Putting the value of f1\(\hat{a} \) € -in (1), we get
f2\(\hat{a} \) € -= 47 – f1
= f2\(\hat{a} \) € -= 47 – 18 = 29
f1\(\hat{a} \) € -= 18, f2\(\hat{a} \) € -= 29
Question. The mean of the given frequency distribution is 62.8, and the sum of frequencies is 50. Calculate the missing frequencies f1\(\hat{a} \) € -and f2: (2013)
| Classes | 0-20 | 20-40 | 40-60 | 60-80 | 80-100 | 100-120 | Total |
| Frequencies | 5 | f1 | 10 | f2 | 7 | 8 | 50 |
Solution. 30+f1+f2=50
50-30-f1=f2
F2=20-f1 (1)
Mean =∑xifi /∑fi\(\hat{a} \) € -=62.8
2060+30f1+70f2/50=62.8
= 2060 + 30f1\(\hat{a} \) € -+ 70 f2\(\hat{a} \) € -= 3140
= 30 f1 + 70 f2\(\hat{a} \) € -= 3140 – 2060 = 1080
= 3 f1 + 7 f2\(\hat{a} \) € -= 108 after dividing by 10
= 3 f1 + 7(20 – f1) = 108 From (1)
=3 f1– 7f1\(\hat{a} \) € -+ 140 = 108
= -4 f1 = -32.
f1 = 8
Putting the value of f1\(\hat{a} \) € -into (i), we get
f2\(\hat{a} \) € -= 20 – 8 = 12
f2\(\hat{a} \) € -= 12
Question. Given that the median for the data is 31, find the value of x and y. (2012)
Solution. Total 40
=22+x+y=40
=X+y=18
=Y=18-x (1)
n/2=40/2=20
median is 31 so median class is 30-40
Median= l + \(\frac{n/2?cf}{f} ×h \)
31= \(\frac{30+20-(11+x)}{18-x}\) x h (as f= y which is equal to 18-x according to statement 1)
= 31- 30 =\(\frac{(20-11-x)}{18-x}\) x 10
= 18 – x = (9 – x)10
= 18 – x = 90 – 10x
=10x -x = 90 – 18
= 9x = 72
= x = 8
Putting the value of x in (1
y = 18 – 8 = 10
x = 8, y = 10
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