Statistics Important Questions

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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) 

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) 

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) 

Solution.

=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) 

Solution.

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) 

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) 

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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