Math Olympiad Test: Statistics- 3 - Free MCQ with solutions Class 10

Mean of 1,3, 4, 5, 7 and 4 is m.
⇒ 1 + 3 + 4 + 5 + 7 + 4 / 6 = m
⇒ m = 4
Now, mean of 3, 2, 2, 4, 3, 3 and p is m - 1
3 + 2 + 2 + 4 + 3 + 3 + p / 7 = 3
⇒ 17 + p = 21 ⇒ p = 4
Arranging 3, 2, 2, 4, 3, 3, and 4 in ascending order, we get 2, 2, 3, 3, 3, 4, 4
∴ Median (q) =   term = 4th term = 3 
⇒ p + q= 4 + 3 = 7

Since, maximum class frequency is 35, so the mode class is 18 - 24.

Now, Mode = 

= 

= 1 8 + 2.22 = 20.22

Mean, median and mode are the measures of central tendency.

Ogive or cumulative frequency curve is used to find the median.

36 x 9 = x₁ + x₂ + x₃ + . . . + x₉ ... (i)
5 x 32 = x₁ + x₂ + x₃ + x₄ + x₅ ,..(ii)
5 x 39 = x₅ + x₆ + x₇ + x₈ + x₉ ...(iii)

Adding (ii) and (iii),

we get 355 = x 1 + x₂ + x ₃ + x₄ + x ₅ + ... + x₉ + X₅

Using (i), we get 355 - 36 x 9 = x₅ 

x₅ = 31

Let the number of agricultural and other workers be n₁ and n₂.
∴ n₁ = 11n₂    ......(i)
Sum of monthly income of agricultural workers = Sn₁ 
Sum of monthly income of other workers = Tn₂
Sum of monthly income of all workers = Sn₁  + Tn₂
∴ Average monthly income of all workers
= 

It is given that,
Mean weight of 34 students = 46.5kg
So the total weight = 34 (46.5) = 1581kg
We know that,
Mean weight of 34 students and teacher = 46.5 + 0.5 = 47kg
So the total weight of 34 students and teacher = 47 (35) = 1645kg
The weight of teacher = Weight of 34 students and teacher – total weight
By substituting the values,
Weight of the teacher = 1645 – 1581 = 64kg
Therefore, the weight of the teacher is 64kg.

 Let the temperatures on first, second, third, fourth and fifth day be x₁, x₂, x₃, x₄ and x₅, respectively
Sum of temperatures of Ifirst four days = 58 * 4

i.e ., x₁ + x₂ + x₃ + x₄ = 232 ...(i)

and sum of temperatures on second, third, fourth and fifth day = 60 x 4

i.e., x₂ + x₃ + x₄ + x₅- = 240 ...(ii) Subtracting (i) from (ii), we get x₅- x₁ = 8 ...(iii)

Also, temperature of first and fifth day were in the ratio 7 : 8. Let the temperatures be 7 k and 8k respectively
From (iii), we have 8k - 7k = 8

⇒ k = 8

Temperature of fifth day = 8 x 8 = 64 degree.

We have given n = 100

⇒ 76 + f₁ + f₂ = 100

⇒  f₁ + f₂ = 24                 ....(i)

Since median is 525, so median class is 500 - 600.

Median = 

⇒ 25 = (14 - f₁) x 5 ⇒ 25 = 70 - 5f₁ 

⇒ 5f₁, = 45

⇒ f₁ = 9

From (i), 9 + f₂ = 24

⇒ f₂ = 24 - 9 = 15

Here, n = 675

⇒ n/2 = 337.5

So, Median class 40 - 50.

Median = 

= 

= 40 + 7.58 = 47.58

Now, maximum frequency is 180.
So, modal class is 40 - 50.

Mode = 

= 

= 48.22