Công cụ giải thống kê

[Analysis]
We have the set S = {2, 4, 6, 8, x, y} with x,y distinct primes in (0,40).
We must check each statement:

I. The maximum possible range = max(S) − min(S).
II. The median of six values is the average of the 3rd and 4th smallest; check parity.
III. With y=37, compare the mean to the median.

[Solution]

I. Maximum possible range
1 The smallest element is $2$.
2 The largest prime <40 is $37$.
3 Thus max range = $37 - 2$ = $35$.
4 Since $35$ > $33$, the maximum possible range is >33.
⇒ Statement I is true.

II. Parity of the median
1 For six data points sorted a₁…a₆ the median = $\frac{a₃ + a₄}{2}$.
2 To show it can be even, pick x,y among small primes: e.g. x=5,y=7 gives S={2,4,5,6,7,8}.
3 Then a₃=5 and a₄=6 so median = $\frac{5 + 6}{2}$ = $5.5$ (not even).
4 But choose x=5,y=7 in another order say S={2,3,5,7,6,8} sorted {2,3,5,6,7,8}, now a₃=5,a₄=6 →5.5 again.
5 To force two odd middle entries use x=7,y=11 giving sorted {2,4,6,7,8,11}, then a₃=6,a₄=7 →6.5.
6 Actually to get an even median we need two odd middles: pick x=5,y=7 and reorder so a₃=5,a₄=7 giving median = $\frac{5 + 7}{2}$ = $6$.
7 Hence the median can be an even integer.
⇒ Statement II is false.

III. Mean versus median when y=37
Let S={2,4,6,8,x,37}.
Case 1 x ≤ 6 gives sorted {2,4,x,6,8,37}. median = $\frac{x + 6}{2}$ mean = $\frac{2 + 4 + 6 + 8 + x + 37}{6}$ = $\frac{57 + x}{6}$ Compare: $\frac{57 + x}{6}$ > $\frac{x + 6}{2}$ ↔ $57 + x$ > $3(x + 6)$ ↔ $57 + x$ > $3x + 18$ ↔ $39$ > $2x$ Since x ≤ 6 implies $2x$ ≤ 12 $39$ > $12$ holds.

Case 2 6 < x ≤ 8 (so x=7): median = $\frac{6 + 7}{2}$ = $6.5$ mean = $\frac{57 + 7}{6}$ ≈ $10.67$ > $6.5$.

Case 3 x > 8 gives sorted {2,4,6,8,x,37}: median = $\frac{6 + 8}{2}$ = $7$ mean = $\frac{57 + x}{6}$ ≥ $\frac{57 + 11}{6}$ = $\frac{68}{6}$ ≈ $11.33$ > $7$.

In every case mean > median.
⇒ Statement III is true.

[Answer]
Statements I and III must be true.

[Analysis]

The problem is a one‐sample z‐test for the mean with known population standard deviation.
We test – Null hypothesis HO: μ = 120 – Alternative hypothesis Ha: μ > 120
at significance level α = 0.05.
Approach: 1. Compute the test statistic z = (x̄ – μ0) / (σ / √n) 2. Find the p‐value for a right‐tailed test: p = P(Z ≥ z_obs) where Z ~ N(0,1)

[Solution]

Step 1 Compute the test statistic $z = \frac{\,x̄ \;-\;\mu_{0}\,}{\,\sigma \,/\,\sqrt{n}\,}$

Step 2 Plug in the numbers $z = \frac{\,105.37 \;-\;120\,}{\,32.17 \,/\,\sqrt{40}\,}$

Step 3 Evaluate numerator and denominator – Numerator: 105.37 – 120 = –14.63 – Denominator: 32.17 ÷ √40 ≈ 32.17 ÷ 6.3249 ≈ 5.087

Step 4 Compute z $z ≈ \frac{\,-14.63\,}{\,5.087\,} ≈ -2.876$

Step 5 Find the p‐value for the right‐tailed test $p\text{-value} = P(Z ≥ -2.876) = 1 \;-\; \Phi(-2.876)$
Using the standard normal table or a calculator: $\Phi(-2.876) ≈ 0.00203$ $p\text{-value} ≈ 1 - 0.00203 = 0.99797$

[Answer]

The p‐value for the test is $p\text{-value} ≈ 0.998$