Q.Assume that height (in cm) of students in a class are as follows [90,102,110,115,85,90,100,110,110]. Mean or average height of the class is

Step 1 — Sum the observations

Step 2 — Set the mean equal to 125

Step 3 — Solve

$x=125-5=120.$ …

Step 1 — Write the mean equation

For the five observations $40,\ k,\ 6k,\ 4k^2,\ 8k-4k^2$:

Step 2 — Simplify the numerator

The $k^2$ terms cancel and the $k$ terms combine:

Step 3 — Solve for k

Why the other options are wrong: $k=15,10,8$ give sums of $265,190,160$ and means of $53,38,32$ — none equal $20$. Only $k=4$ satisfies the equation. …

Step 1 — Find the total sum

Mean $= 28$ over $5$ observations, so

Step 2 — Subtract the known observations

The three given values total $10 + 23 + 62 = 95$, so the remaining two satisfy

Step 3 — Use the difference and solve

Given $x - y = 13$, add the two equations:

Check: $29 + 16 = 45$ ✓ and $29 - 16 = 13$ ✓.

Why the other options are wrong: (A) $30+17=47$, (B) $38+15=53$, (D) $35+22=57$ — none sum to the required $45$, so only $29$ and $16$ fit both conditions. …

Step 1 — State the property

For observations $x_1,\ldots,x_n$ with mean $\bar{x}$:

Step 2 — Show why

Expand the sum:

Positive deviations (values above the mean) exactly offset negative deviations (values below it).

Step 3 — Reject the false statements

• Sum of deviations is zero, not always positive → (A) is false.
• The mean is affected by extreme values → (B) is false.
• The mean can be computed for grouped data using $\dfrac{\sum f x}{\sum f}$ → (D) is false.

So only (C) is a genuine property. …

Step 1 — Recover the original total

Mean $=\dfrac{\text{sum}}{n}$, so sum $=$ mean $\times n = 15\times 10 = 150$.

Step 2 — Adjust the total for the replacement

Remove 12 and add 22:

Step 3 — Recompute the mean

Why the other options are wrong

• (B) 14 wrongly subtracts the change instead of adding it. …

Step 1 — Set up the formula

X are the values and Y are their frequencies (f), so the arithmetic mean of the frequency distribution is:

Step 2 — Compute $\sum f$ and $\sum fX$

Step 3 — Divide

$\bar{X}=\dfrac{299}{73}=4.0959\approx 4.10$ …

Step 1 — Find the total from the mean

Step 2 — Isolate the remaining two numbers

The biggest number is $180$, so the other two add to

Step 3 — Split using the difference

Let the two numbers be $a$ and $b$ with $a+b=225$ and $a-b=25$.

Both are below the biggest value $180$, so the data is consistent and the smallest number is $100$. …

Step 1 — Write the sum of deviations

Step 2 — Substitute the mean

Since $\bar{x}=\dfrac{\sum x_i}{n}$, we have $n\bar{x}=\sum x_i$, so

This is a defining property of the arithmetic mean — it is the point of balance of the data.

Why the other options are wrong: $\dfrac{\sum x_i}{n}$ (A) is the mean itself, (B) is the sample standard deviation, and (C) is a mis-formed average — none equals the sum of deviations. …

Step 1 — Build Σf and Σfx

Step 2 — Apply the mean formula

Step 3 — Solve

$3Y - 2.6Y = 62.4 - 56 \;\Rightarrow\; 0.4Y = 6.4 \;\Rightarrow\; Y = 16.$ …

Step 1 — Total of all seven

Step 2 — Totals of the two overlapping groups

These two groups name $4 + 4 = 8$ positions, but there are only 7 numbers, so exactly one number belongs to both groups (the 4th, or middle, number).

Step 3 — Find the shared number

By inclusion–exclusion, sum of the two groups counts the shared number twice:

$\text{shared number} = 104 - 84 = 20$ …