the frequency distribution table shows the number of mango trees in a grove and their yield of mangoes. find the median of data.
(no.of mangoes) (50-100)(100-150) (150-200)(200-250)(250-300)
(no.of trees) (33) (30) (90) (80) (17)
please tell how to solve
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Page No 138:
Question 1:
The following table shows the number of students and the time they utilized daily for their studies. Find the mean time spent by students for their studies by direct method.
Time (hrs.)0 - 22 - 44 - 66 - 88 - 10No. of students71812103
Answer:
Class
(Time in hours)Class Mark
xiFrequency
(Number of students)
fiClass mark × Frequency
xifi0 - 21772 - 4318544 - 6512606 - 8710708 - 109327 ∑fi=50∑xifi=218
Mean = ∑xifi∑fi
=21850
= 4.36 hours
Hence, the mean time spent by students for their studies is 4.36 hours.
Page No 138:
Question 2:
In the following table, the toll paid by drivers and the number of vehicles is shown. Find the mean of the toll by 'assumed mean' method.
Toll (Rupees)300 - 400400 - 500500 - 600600 - 700700 - 800No. of vehicles801101207040
Answer:
Class
(Toll in rupees)Class Mark
xidi = xi − AFrequency
(Number of vehicles
fiFrequency × deviation
fi × di300 - 400 350−20080−16000400 - 500450−100110−11000500 - 600550 = A01200600 - 700650100707000700 - 800750200408000 ∑fi=420∑fidi=-12000
Required Mean = A+∑fidi∑fi
=550-12000420
= 550 − 28.57
= Rs 521.43
Hence, the mean of toll is Rs 521.43.
Page No 138:
Question 3:
A milk centre sold milk to 50 customers. The table below gives the number of customers and the milk they purchased. Find the mean of the milk sold by direct method.
Milk Sold (Litre)1 - 22 - 33 - 44 - 55 - 6No. of Customers17131073
Answer:
Class
(Milk sold in litres)Class Mark
xiFrequency
(Number of customers)
fiClass mark × Frequency
xifi1 - 21.51725.52 - 32.51332.53 - 43.510354 - 54.5731.5 5 - 65.5316.5 ∑fi=50∑xifi=141
Mean = ∑xifi∑fi
=14150
= 2.82 litres
Hence, the mean of the milk sold is 2.82 litres.
Page No 138:
Question 4:
A frequency distribution table for the production of oranges of some farm owners is given below. Find the mean production of oranges by 'assumed mean' method.
Production
(Thousand rupees)
25 - 3030 - 3535 - 4040 - 4545 - 50No. of Customers2025151010
Answer:
Class
(Production in
Thousand rupees)
Class Mark
xidi = xi − AFrequency
(Number of farm owners)
fiFrequency × deviation
fi × di25 - 30 27.5−1020−20030 - 3532.5−525−12535- 4037.5= A015040 - 4542.55105045 - 5047.51010100 ∑fi=80∑fidi=-175
Required Mean = A+∑fidi∑fi
37.5-17580
= 37.5 − 2.19
= 35.31 thousand rupees
= Rs 35310
Hence, the mean production of oranges is Rs 35310.
Page No 138:
Question 5:
A frequency distribution of funds collected by 120 workers in a company for the drought affected people are given in the following table. Find the mean of the funds by 'step deviation' method.
Fund (Rupees)
0 - 500500 - 10001000 - 15001500 - 20002000 - 2500No. of workers3528321530
Answer:
Class
(Production in
Thousand rupees)
Class Mark
xidi = xi − Aui=dihFrequency
(Number of farm owners)
fiFrequency × deviation
fi × ui0 - 500 250−1000−235−70500 - 1000 750−500−128−281000 - 1500 1250 = A003201500 - 2000 1750500115152000 - 2500 2250100021020 ∑fi=120∑fiui=-63
Required Mean = A+h∑fiui∑fi
=1250-63120500
= 1250 − 262.5
= Rs 987.5
Hence, the mean of the funds is Rs 987.5.
Page No 138:
Question 6:
The following table gives the information of frequency distribution of weekly wages of 150 workers of a company. Find the mean of the weekly wages by 'step deviation' method.
Weekly wages (Rupees)
1000 - 20002000 - 30003000 - 40004000 - 5000No. of workers25455030
Answer:
Class
(Weekely wages rupees)
Class Mark
xidi = xi − Aui=dihFrequency
(Number of workers)
fiFrequency × deviation
fi × ui1000 - 2000 1500−2000−225−502000 - 3000 2500−100−145−453000 - 4000 3500 = A005004000 - 5000 4500100013030 ∑fi=150∑fiui=-65
Required Mean = A+h∑fiui∑fi
=3500-651501000
= 3500 − 433.33
= Rs 3066.67
Hence, the mean of the weekly wages is Rs 3066.67.
Page No 138:
Question 1:
The following table shows the number of students and the time they utilized daily for their studies. Find the mean time spent by students for their studies by direct method.
Time (hrs.)0 - 22 - 44 - 66 - 88 - 10No. of students71812103
Answer:
Class
(Time in hours)Class Mark
xiFrequency
(Number of students)
fiClass mark × Frequency
xifi0 - 21772 - 4318544 - 6512606 - 8710708 - 109327 ∑fi=50∑xifi=218
Mean = ∑xifi∑fi
=21850
= 4.36 hours
Hence, the mean time spent by students for their studies is 4.36 hours.
Page No 138:
Question 2:
In the following table, the toll paid by drivers and the number of vehicles is shown. Find the mean of the toll by 'assumed mean' method.
Toll (Rupees)300 - 400400 - 500500 - 600600 - 700700 - 800No. of vehicles801101207040
Answer:
Class
(Toll in rupees)Class Mark
xidi = xi − AFrequency
(Number of vehicles
fiFrequency × deviation
fi × di300 - 400 350−20080−16000400 - 500450−100110−11000500 - 600550 = A01200600 - 700650100707000700 - 800750200408000 ∑fi=420∑fidi=-12000
Required Mean = A+∑fidi∑fi
=550-12000420
= 550 − 28.57
= Rs 521.43
Hence, the mean of toll is Rs 521.43.
Page No 138:
Question 3:
A milk centre sold milk to 50 customers. The table below gives the number of customers and the milk they purchased. Find the mean of the milk sold by direct method.
Milk Sold (Litre)1 - 22 - 33 - 44 - 55 - 6No. of Customers17131073
Answer:
Class
(Milk sold in litres)Class Mark
xiFrequency
(Number of customers)
fiClass mark × Frequency
xifi1 - 21.51725.52 - 32.51332.53 - 43.510354 - 54.5731.5 5 - 65.5316.5 ∑fi=50∑xifi=141
Mean = ∑xifi∑fi
=14150
= 2.82 litres
Hence, the mean of the milk sold is 2.82 litres.
Page No 138:
Question 4:
A frequency distribution table for the production of oranges of some farm owners is given below. Find the mean production of oranges by 'assumed mean' method.
Production
(Thousand rupees)
25 - 3030 - 3535 - 4040 - 4545 - 50No. of Customers2025151010
Answer:
Class
(Production in
Thousand rupees)
Class Mark
xidi = xi − AFrequency
(Number of farm owners)
fiFrequency × deviation
fi × di25 - 30 27.5−1020−20030 - 3532.5−525−12535- 4037.5= A015040 - 4542.55105045 - 5047.51010100 ∑fi=80∑fidi=-175
Required Mean = A+∑fidi∑fi
37.5-17580
= 37.5 − 2.19
= 35.31 thousand rupees
= Rs 35310
Hence, the mean production of oranges is Rs 35310.
Page No 138:
Question 5:
A frequency distribution of funds collected by 120 workers in a company for the drought affected people are given in the following table. Find the mean of the funds by 'step deviation' method.
Fund (Rupees)
0 - 500500 - 10001000 - 15001500 - 20002000 - 2500No. of workers3528321530
Answer:
Class
(Production in
Thousand rupees)
Class Mark
xidi = xi − Aui=dihFrequency
(Number of farm owners)
fiFrequency × deviation
fi × ui0 - 500 250−1000−235−70500 - 1000 750−500−128−281000 - 1500 1250 = A003201500 - 2000 1750500115152000 - 2500 2250100021020 ∑fi=120∑fiui=-63
Required Mean = A+h∑fiui∑fi
=1250-63120500
= 1250 − 262.5
= Rs 987.5
Hence, the mean of the funds is Rs 987.5.
Page No 138:
Question 6:
The following table gives the information of frequency distribution of weekly wages of 150 workers of a company. Find the mean of the weekly wages by 'step deviation' method.
Weekly wages (Rupees)
1000 - 20002000 - 30003000 - 40004000 - 5000No. of workers25455030
Answer:
Class
(Weekely wages rupees)
Class Mark
xidi = xi − Aui=dihFrequency
(Number of workers)
fiFrequency × deviation
fi × ui1000 - 2000 1500−2000−225−502000 - 3000 2500−100−145−453000 - 4000 3500 = A005004000 - 5000 4500100013030 ∑fi=150∑fiui=-65
Required Mean = A+h∑fiui∑fi
=3500-651501000
= 3500 − 433.33
= Rs 3066.67
Hence, the mean of the weekly wages is Rs 3066.67.
Answered by
7
Answer:
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sayali29:
only last answer 184.4 is given
In which class book.is it 10th
ssc board
10th
algebra
Okk thanks I will check
ok thank you
bye
did look it or not
answer is wrong
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