Q.Distinguish between Karl Pearson's coefficient of correlation and Spearman's rank correlation coefficient, and state one situation in which the rank method is preferable.

Step 1 — find the original Σd².

With $n = 10$, $n(n^2-1) = 10 \times 99 = 990$ and $R = 0.5$:

Step 2 — correct the wrong term.

One difference was recorded as 3 instead of 7. Remove $3^2 = 9$ and add $7^2 = 49$:

Step 3 — recompute R. …

Step 1 — Write Spearman's rank correlation formula

Step 2 — Isolate the fraction

With $r=\frac{75}{99}$ and $\sum d^2=40$:

Step 3 — Solve for n

Since $n(n^2-1)=990$ and $10\times(100-1)=10\times 99=990$, we get $n=10$. …

Step 1 — Write the rank-correlation formula

Step 2 — Substitute $n=10,\ \Sigma d^2=5$

Why the other options are wrong: $0.95,0.96,0.99$ come from rounding slips or a wrong $n(n^2-1)$. …

Step 1 — Formula

Step 2 — Substitute $n=9,\ \Sigma d^2=62$

Step 3 — Evaluate

Why the other options are wrong: (A) 1 ignores the correction term; (C) 0.52 is $\dfrac{372}{720}$ (the term itself, not $1-$ it); (D) 0.87 uses a wrong denominator. …

Step 1 — Get r from the given relation

$\Sigma d^2 = 60$ is 120 times $r$, so

Step 2 — Spearman's rank correlation formula

Substitute $r=0.5$, $\Sigma d^2=60$:

Step 3 — Solve for n

Testing $n=9$: $9(81-1)=9\times80=720$. ✓ …

Step 1 — Spearman's formula

Step 2 — Substitute n = 10, r = 0.8

Step 3 — Solve for ∑d²

$\frac{6\,\Sigma d^{2}}{990}=0.2 \;\Rightarrow\; 6\,\Sigma d^{2}=198 \;\Rightarrow\; \Sigma d^{2}=33$ …

Step 1 — State the standard formula

Step 2 — Match the option

The correct form subtracts the term and uses n(n²−1) in the denominator — option (D).

Step 1 — Convert scores to ranks (1 = lowest)

Maths 3,5,8,4,7,10 → ranks 1,3,5,2,4,6.

Statistics 6,4,9,8,1,2 → ranks 4,3,6,5,1,2.

Step 2 — Differences and their squares

Step 3 — Apply the formula

$r_s = 1 - \frac{6\sum d^2}{n(n^2-1)} = 1 - \frac{6(44)}{6(35)} = 1 - 1.257 = -0.257$ …