Skip to content
Question of 90

Q.Find the mean deviation about the median for the following distribution. Marks: 10-20,20-30,30-40,40-50,50-60,60-7010\text{-}20, 20\text{-}30, 30\text{-}40, 40\text{-}50, 50\text{-}60, 60\text{-}70; No. of students: 3,4,7,8,2,13, 4, 7, 8, 2, 1.

Rajasthan RbseRajasthan Board Senior Secondary Part-I Examination 2018Subjective· 4mImportance★★★★★
0% · 0/90 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Find the median class and median, then compute 1N∑fi∣xi−M∣\dfrac{1}{N}\sum f_i |x_i - M| using class midpoints.

Data:

Class10-2020-3030-4040-5050-6060-70
ff347821

N=3+4+7+8+2+1=25N = 3+4+7+8+2+1 = 25

Step 1: Find the median.

Cumulative frequencies: 3,7,14,22,24,253, 7, 14, 22, 24, 25.

N/2=12.5N/2 = 12.5, and the cumulative frequency first exceeds this in the class 30-4030\text{-}40 (cf just before it =7=7, f=7f=7, h=10h=10, l=30l=30).

M=l+N/2−cff×h=30+12.5−77×10=30+557≈37.86M = l + \dfrac{N/2 - cf}{f} \times h = 30 + \dfrac{12.5-7}{7}\times10 = 30 + \dfrac{55}{7} \approx 37.86

Step 2: Compute ∣xi−M∣|x_i - M| for each midpoint xix_i.

Midpoints: 15,25,35,45,55,6515, 25, 35, 45, 55, 65

| xix_i | fif_i | ∣xi−37.86∣|x_i - 37.86| | fi∣xi−M∣f_i|x_i-M| |

|---|---|---|---| …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.