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**Short Answer Type Questions**

**Q.1. The following data gives the information on the observed lifetimes (in hours) of 225 electrical components:Determine the modal lifetimes of the components. [CBSE 2019 (30/5/1)]Ans.** Here, the maximum class frequency is 61 and the class corresponding to this frequency is 60-80. So, the modal class is 60-80.

Here, l = 60, h = 20 ,f

∴

Hence, modal lifetime of the components is 65.625 hours

Therefore, median salary = 13.42 (in thousand Rs.)

**Long Answer Type Questions**

**Q.1. If the median of the following frequency distribution is 32.5. Find the values of f _{1} and f_{2}. [Delhi 2019]**

Ans.

⇒

⇒

⇒

⇒ 15 = 5(6 - f

Ans.

Let assumed mean A = 70 and class size h = 10

Now, we have

∴ **Q.3. The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs 18. Find the missing frequency f. [CBSE 2019 (30/1/2)]Ans.** Let the assumed mean A = 16 and class size h = 2, here we apply step deviation method.

Now, we have,

We have, mean = 18, A = 16 and h = 2

⇒ f + 44 = 2f + 24

⇒ f = 44 - 24

⇒ f = 20

Hence, the missing frequency is 20.

Ans.

N = 100

N/2=50

∴ Median = 34.34 = 34.3

Ans.

⇒ 18(40+f) = 704+20f

⇒ 720+18f - 20f + 704

⇒ 720-704 = 20f - 18f

Convert the distribution above to a less than type cumulative frequency distribution and draw its ogive.

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