| Scores | 1 | 2 | 3 | 4 | 5 | 6 |
| Frequency | 2 | 5 | 13 | 11 | 9 | 10 |
The table shows the distribution of outcomes when a die is thrown 50 times. Calculate the :
(a) Mean deviation of the distribution ; (b) probability that a score selected at random is at least a 4.
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Correct Answer: Option n
Explanation:
(a)
\(Mean (\bar{x}) = \frac{\sum fx}{\sum f} = \frac{200}{50} = 4\)
Hence, Mean Deviation = \(\frac{\sum f|d|}{\sum f} = \frac{69}{50} \)
= \(1.38\)
(b) Let E denote the event of getting a score of at least 4.
\(n(E) = 11 + 9 + 10 = 30\)
\(p(E) = \frac{n(E)}{n(S)} = \frac{30}{50}\)
= \(0.6\)
(a)
| Scores (x) | Frequency (f) | \(fx\) |
| 1 | 2 | 2 |
| 2 | 5 | 10 |
| 3 | 13 | 39 |
| 4 | 11 | 44 |
| 5 | 9 | 45 |
| 6 | 10 | 60 |
| \(\sum f = 50\) | \(\sum fx = 200\) |
\(Mean (\bar{x}) = \frac{\sum fx}{\sum f} = \frac{200}{50} = 4\)
| \(d = x - 4\) | \(|d|\) | f | \(f|d|\) |
| -3 | 3 | 2 | 6 |
| -2 | 2 | 5 | 10 |
| -1 | 1 | 13 | 13 |
| 0 | 0 | 11 | 0 |
| 1 | 1 | 9 | 9 |
| 2 | 2 | 10 | 20 |
| \(\sum f|d| = 69\) |
Hence, Mean Deviation = \(\frac{\sum f|d|}{\sum f} = \frac{69}{50} \)
= \(1.38\)
(b) Let E denote the event of getting a score of at least 4.
\(n(E) = 11 + 9 + 10 = 30\)
\(p(E) = \frac{n(E)}{n(S)} = \frac{30}{50}\)
= \(0.6\)