Tag: mean

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Multiple choice statistics measures of central tendency geometric and harmonic mean geometric mean mean

The mean of the following frequency distribution is 62.8 and the sum of all the frequencies is 50. Compute the missing frequency $\displaystyle f _{1}$ and $\displaystyle f _{2}$.

Class 0-20 20-40 40-60 60-80 80-100 100-120
Frequency 5 $\displaystyle f _{1}$ 10 $\displaystyle f _{2}$ 7 8
  1. $5, 8$
  2. $6, 12$
  3. $8, 11$
  4. $8, 12$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Class       Frequency(f)  ClassMark (x)         fx
0-20         5      10         50
20-40     ${f} _{1} $      30       $ 30{f} _{1} $
40-60         10       50          500
60-80      ${f} _{2} $        70        $ 70{f} _{2} $
80-100          7         90          630
100-120           8         110           880
Total $30 + {f} _{1} +{f} _{2} $   $ 2060  + 30{f} _{1}+70{f} _{2} $

Given $ 30 + {f} _{1} +{f} _{2} = 50 $
$ => {f} _{1} +{f} _{2} = 20 $   -- (1)

Given, Mean $ = \cfrac { \sum { fx }  }{ \sum { f }} =62.8 $
$ => \cfrac { 2060  + 30{f} _{1}+70{f} _{2}}{30 + {f} _{1} +{f} _{2}} = 62.8 $

$ =>  2060  + 30{f} _{1}+70{f} _{2} = 1884 + 62.8{f} _{1} + 62.8{f} _{2} $ 

$ 32.8{f} _{1} - 7.2{f} _{2} =176 $

=> $ 8.2{f} _{1} - 1.8{f} _{2} = 44 $

=> $ 4.1{f} _{1} - 0.9{f} _{2} = 22 $ -- (2)

Solving both equations 1, 2, we get
$ {f} _{1} = 8, {f} _{2} = 12 $

Multiple choice mean and median mean maths assumed mean method assumed mean method of finding mean measure of central tendency

What is the value of mean for the following data:

Marks No. of Student
$5-14$ $10$
$15-24$ $18$
$25-34$ $32$
$35-44$ $26$
$45-54$ $14$
$55-64$ $10$
  1. $30$
  2. $29$
  3. $33.68$
  4. $34.21$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

To find the mean, use the formula (sum of f*x) / N. The midpoints (x) are 9.5, 19.5, 29.5, 39.5, 49.5, 59.5. Summing f*x gives 3368, and N is 100, resulting in 33.68.

Multiple choice mean and median mean maths assumed mean method assumed mean method of finding mean measure of central tendency

Mr. R purchases $20\ kgs$. of Wheat, $10\ kgs.$ of rice and $2\ kgs.$ of ghee every month. If the price of wheat is $Rs. 10$ per kg., price of rice is $Rs. 14$ per kg. and price of ghee is $Rs. 120$ per kg. Find the average price per kg. per month. Arithmetic mean $=$?

  1. $48$
  2. $18.125$
  3. $48.125$
  4. $18$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The average price per kg is the total cost divided by the total weight. Total cost = (20*10) + (10*14) + (2*120) = 200 + 140 + 240 = 580. Total weight = 20 + 10 + 2 = 32. Average = 580 / 32 = 18.125.

Multiple choice mean and median mean maths assumed mean method assumed mean method of finding mean measure of central tendency

If there are two groups containing $30$ and $20$ observations and having $50$ and $60$ as arithmetic means, then the combined arithmetic mean is ________.

  1. $55$
  2. $56$
  3. $54$
  4. $52$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Combined mean = (n1*x1 + n2*x2) / (n1 + n2). Calculation: (30*50 + 20*60) / (30 + 20) = (1500 + 1200) / 50 = 2700 / 50 = 54.

Multiple choice mean and median mean maths assumed mean method assumed mean method of finding mean measure of central tendency

What is the value of mean for the following data.

Class interval Frequency
$350-369$ $15$
$370-389$ $27$
$390-409$ $31$
$410-429$ $19$
$430-449$ $13$
$450-469$ $6$
  1. $400$
  2. $400.58$
  3. $394$
  4. $394.50$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Class Intervals Mid-values(x) Frequency(f) fx
$350-369$ $359.5$ $15$ $5,392.5$
$370.389$ $379.5$ $27$ $10,246.5$
$390-409$ $399.5$ $31$ $12,384.5$
$410-429$ $419.5$ $19$ $7,970.5$
$430-449$ $439.5$ $13$ $5,713.5$
$450-469$ $459.5$ $6$ $2,757$
$\displaystyle\sum f=111$ $\displaystyle\sum fx=44,464.5$

Arithmetic mean $=44,464.5/111=400.58$.

Multiple choice mean and median mean maths assumed mean method assumed mean method of finding mean measure of central tendency

The following is the distribution of weekly wages of workers in a factory. Calculate the arithmetic mean of the distribution.

Weekly Wages (Rs.) No. of Workers
$240-269$ $7$
$270-299$ $19$
$300-329$ $27$
$330-359$ $15$
$360-389$ $12$
$390-419$ $12$
$420-449$ $8$
  1. $352.3$
  2. $344.5$
  3. $226.7$
  4. $336.7$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Class Intervals Mid-values (x) Frequency(f) fx
$240-269$ $254.5$ $7$ $1,781.5$
$270-299$ $284.5$ $19$ $5,405.5$
$300-329$ $314.5$ $27$ $8,491.5$
$330-359$ $344.5$ $15$ $5,167.5$
$360-389$ $374.5$ $12$ $4,494$
$390-419$ $404.5$ $12$ $4,854$
$420-449$ $434.5$ $8$ $3,476$
$\displaystyle\sum f=100$ $\displaystyle\sum fx=33,670$

Arithmetic mean $=33,670/100=336.7$.

Multiple choice mean and median mean maths assumed mean method assumed mean method of finding mean measure of central tendency

The mean weight of $98$ students is found to be $50$ lbs. It is later discovered that the frequency of the class interval $(30-40)$ was wrongly taken as $8$ instead of $10$. Calculate the correct mean.

  1. $49.00$
  2. $49.50$
  3. $49.25$
  4. $49.70$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Incorrect mean,
$X=50$kg.
$\displaystyle\sum f _i=98$
Incorrect X$=\displaystyle\frac{Incorrect \displaystyle\sum f _iX _i}{\displaystyle\sum f _i}$
$50=\displaystyle\frac{Incorrect \displaystyle\sum f _iX _i}{98}$
$\therefore$ Incorrect $\displaystyle\sum f _iX _i=98\times 50=4900$
Now, Correct $\displaystyle\sum f _iX _i=$Incorrect $\displaystyle\sum f _iX _i-(8\times 35)+(10\times 35)$
Note, the class-mark of class interval $(30-40)$ is $35$ and for the calculation of the mean, we consider class marks.
Correct $\displaystyle\sum f _iX _i=4900-280+350$
$=4,970$
Also, Correct $\displaystyle\sum f _i=98+2=100$
$\therefore$ Correct Mean$=\displaystyle\frac{Correct \displaystyle\sum f _iX _i}{Correct \displaystyle\sum f _i}$
$=\displaystyle\frac{4970}{100}$
$X=49.70$lbs.

Multiple choice mean and median mean maths assumed mean method assumed mean method of finding mean measure of central tendency

Consider the following frequency distribution.

Class Intervals $0-10$ $10-20$ $20-30$ $30-40$
Frequency $8$ $10$ $12$ $15$

Arithmetic mean $=$?

  1. $39.65$
  2. $22.55$
  3. $32.55$
  4. $23.56$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Class Intervals Mid-values(x) Frequency(f) fx
$01-0$ $5$ $8$ $40$
$10-20$ $15$ $10$ $150$
$20-30$ $25$ $12$ $300$
$30-40$ $35$ $15$ $525$
$\displaystyle\sum f=45$ $\displaystyle\sum fx =1,015$

Arithmetic mean $=1,015/45=22.55$.

Multiple choice mean and median mean maths assumed mean method assumed mean method of finding mean measure of central tendency

Coefficient of variation of a distribution is $60$ and its standard deviation is $21$, then its arithmetic mean is?

  1. $36$
  2. $37$
  3. $35$
  4. $38$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The coefficient of variation (CV) is calculated as (Standard Deviation / Mean) * 100. Rearranging the formula gives Mean = (Standard Deviation / CV) * 100. Substituting the given values: Mean = (21 / 60) * 100 = 0.35 * 100 = 35.

Multiple choice mean and median mean maths assumed mean method assumed mean method of finding mean measure of central tendency

Arithmetic mean for grouped data can be calculated by _________.

  1. direct method

  2. assumed mean method

  3. step deviation method

  4. all of the above
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Arithmetic mean refers to the average amount in a given group of data. There are many ways to calculate arithmetic mean like direct method where all the data are added up and then divided by the number of figures in the data in order to ascertain the mean class or assumed mean method and step deviation method, the data of the given class is reduced into smaller units which makes it easy to do calculation and ascertain the mean of the class.