(a) The universal set U is the set of integers, P, Q and R are subsets of U defined as follows:
\(P = x : x \leq 2 \) ; \(Q = x : -7 < x < 15\) ; \(R = x : -2 \leq x < 19\).
Find (i) \(P \cap Q\) ; (ii) \(P \cap (Q \cup R')\), where R' is the complement of R with respect to U.
(b) The following data shows the marks of 40 students in a History examination.
41 52 37 56 63 48 65 46 54 32 51 66 74 23 35 61 58 44 49 53 45 57 56 38 59 28 50 49 67 56 36 45 79 68 43 56 26 47 55 71.
(i) Form a grouped frequency table with the class intervals 20 - 29, 30 - 39, 40 - 49 etc; (ii) Find the mean of the distribution.
(a) With U the set of integers: \(P = \{x : x \le 2\}\), \(Q = \{x : -7 < x < 15\}\), \(R = \{x : -2 \le x < 19\}\).
(i) \(P \cap Q\) contains integers that are both \(\le 2\) and strictly between \(-7\) and \(15\), i.e. \(-6 \le x \le 2\):
\[ P \cap Q = \{-6, -5, -4, -3, -2, -1, 0, 1, 2\} = \{x : -6 \le x \le 2\}. \]
(ii) First find \(R' = \{x : x < -2\ \text{or}\ x \ge 19\} = \{x : x \le -3\ \text{or}\ x \ge 19\}\).
Then \(Q \cup R'\): \(Q\) gives \(-6 \le x \le 14\) and \(R'\) gives all \(x \le -3\) together with \(x \ge 19\); their union is \(\{x : x \le 14\} \cup \{x : x \ge 19\}\).
Now intersect with \(P = \{x : x \le 2\}\). Every integer \(\le 2\) is already \(\le 14\), so
\[ P \cap (Q \cup R') = \{x : x \le 2\} = P. \]
(b)(i) Grouped frequency table (40 marks):
| Class | Frequency \(f\) | Midpoint \(x\) |
|---|
| 20 - 29 | 3 | 24.5 |
| 30 - 39 | 5 | 34.5 |
| 40 - 49 | 10 | 44.5 |
| 50 - 59 | 13 | 54.5 |
| 60 - 69 | 6 | 64.5 |
| 70 - 79 | 3 | 74.5 |
| Total | 40 | |
(ii) Mean: \(\sum fx = 3(24.5)+5(34.5)+10(44.5)+13(54.5)+6(64.5)+3(74.5) = 2010\).
\[ \bar{x} = \frac{\sum fx}{\sum f} = \frac{2010}{40} = 50.25 \approx 50.3. \]
(a) With U the set of integers: \(P = \{x : x \le 2\}\), \(Q = \{x : -7 < x < 15\}\), \(R = \{x : -2 \le x < 19\}\).
(i) \(P \cap Q\) contains integers that are both \(\le 2\) and strictly between \(-7\) and \(15\), i.e. \(-6 \le x \le 2\):
\[ P \cap Q = \{-6, -5, -4, -3, -2, -1, 0, 1, 2\} = \{x : -6 \le x \le 2\}. \]
(ii) First find \(R' = \{x : x < -2\ \text{or}\ x \ge 19\} = \{x : x \le -3\ \text{or}\ x \ge 19\}\).
Then \(Q \cup R'\): \(Q\) gives \(-6 \le x \le 14\) and \(R'\) gives all \(x \le -3\) together with \(x \ge 19\); their union is \(\{x : x \le 14\} \cup \{x : x \ge 19\}\).
Now intersect with \(P = \{x : x \le 2\}\). Every integer \(\le 2\) is already \(\le 14\), so
\[ P \cap (Q \cup R') = \{x : x \le 2\} = P. \]
(b)(i) Grouped frequency table (40 marks):
| Class | Frequency \(f\) | Midpoint \(x\) |
|---|
| 20 - 29 | 3 | 24.5 |
| 30 - 39 | 5 | 34.5 |
| 40 - 49 | 10 | 44.5 |
| 50 - 59 | 13 | 54.5 |
| 60 - 69 | 6 | 64.5 |
| 70 - 79 | 3 | 74.5 |
| Total | 40 | |
(ii) Mean: \(\sum fx = 3(24.5)+5(34.5)+10(44.5)+13(54.5)+6(64.5)+3(74.5) = 2010\).
\[ \bar{x} = \frac{\sum fx}{\sum f} = \frac{2010}{40} = 50.25 \approx 50.3. \]
