Skip to content
NCERT Exemplar · Q13

Q.Mean and standard deviation of 100 items are 50 and 4, respectively. Find the sum of all the item and the sum of the squares of the items.

Yanam BieapShort· 2mImportance★★★★★est
63% · 57/90 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 →

Use the definitions of mean and standard deviation to work backwards: mean gives the sum directly, and the variance formula (which involves standard deviation) gives the sum of squares. Sum = 5000, Sum of squares = 251600.

The mean tells us the average value of all items, while the standard deviation measures how spread out the values are from that average. When we know these summary statistics, we can reverse-engineer the raw sums that went into computing them. The key is to recall the formulas that define these measures and solve for the quantities we want.

The mean of nn items is defined as:

xˉ=∑xin\bar{x} = \frac{\sum x_i}{n}

The standard deviation is the square root of the variance, and variance has the computational formula:

σ2=∑xi2n−xˉ2\sigma^2 = \frac{\sum x_i^2}{n} - \bar{x}^2

This second formula is particularly useful because it directly relates the sum of squares to quantities we know.

Finding the sum of all items

  1. Apply the mean formula. We know xˉ=50\bar{x} = 50 and n=100n = 100:

50=∑xi10050 = \frac{\sum x_i}{100}

  1. Solve for the sum. Multiply both sides by 100:

∑xi=50×100=5000\sum x_i = 50 \times 100 = 5000

That's straightforward—the mean is just the total divided by the count, so the total is mean times count.

Finding the sum of squares

  1. Start with the variance formula. The standard deviation is σ=4\sigma = 4, so the variance is: σ2=42=16\sigma^2 = 4^2 = 16 …

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.