S3 June 2013 Q4
4. Customers at a post office are timed to see how long they wait until being served at the counter. A random sample of 50 customers is chosen and their waiting times, \(x\) minutes, are summarised in Table 1.
| Waiting time in minutes (\(x\)) | Frequency |
|---|---|
| 0–3 | 8 |
| 3–5 | 12 |
| 5–6 | 13 |
| 6–8 | 9 |
| 8–12 | 8 |
Table 1
The post office manager believes that the customers’ waiting times can be modelled by a normal distribution.
Assuming the data is normally distributed, she calculates the expected frequencies for these data and some of these frequencies are shown in Table 2.
| Waiting Time | \(x \lt 3\) | 3–5 | 5–6 | 6–8 | \(x \gt 8\) |
| Expected Frequency | 8.56 | 12.73 | 7.56 | \(a\) | \(b\) |
Table 2
| Scheme | Marks |
|---|---|
| \(\bar{x} = \dfrac{8 \times 1.5 + 12 \times 4 + 13 \times 5.5 + 9 \times 7 + 8 \times 10}{50} = \dfrac{274.5}{50} = 5.49\) (*) | B1cso |
| \(s^2 = \dfrac{8 \times 1.5^2 + 12 \times 4^2 + 13 \times 5.5^2 + 9 \times 7^2 + 8 \times 10^2}{49} - \dfrac{50}{49}5.49^2, = 6.88\) (*) | M1,A1cso |
| (3) |
Notes
B1cso for denominator of 50 and at least 3 products on num or 274.5 on num
M1 for a correct expression with at least 3 correct products on num or \(\dfrac{1844.25}{49} - \dfrac{1507.005}{49}\) or \(\dfrac{337.245}{49}\) or \(\left(\dfrac{7377}{200} - 5.49^2\right) \times \dfrac{50}{49}\) etc Allow 3sf accuracy
A1cso for 6.88 with M1 scored and no incorrect working seen
| Scheme | Marks |
|---|---|
| \(a = \quad 50 \times \mathrm{P}(6 \lt X \lt 8) = 50 \times \mathrm{P}(0.194.. \lt Z \lt 0.956..)\) | M1 |
| \(a = 12.81\) (tables) or 12.68 (calc) | A1 |
| \(b = 50 - (28.85 + a) \quad = 8.34\) (tables) or 8.47 (calc) | A1ft |
| (3) |
Notes
M1 a full method for \(a\) or \(b\) using the normal dist. Correct use of (6), 8, 5.49 and \(\sqrt{6.88}\) seen
1st A1 for \(a\) in range 12.68 ~ 12.81 or \(b\) in range 8.34~ 8.47 or awrt these values
2nd A1ft for 50 – 28.85 – their \(a\) (or \(b\)) (but requires M1). Allow awrt 3sf. Must add up to 50
| Scheme | Marks | ||||||||||||||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| \(\mathrm{H}_0\): Normal distribution is a good fit \(\mathrm{H}_1\): Normal distribution is not a good fit | B1 | ||||||||||||||||||||||||||||||
| M1 A1 | ||||||||||||||||||||||||||||||
| \(\displaystyle\sum\frac{O^2}{E} - N = 5.087\ldots \sim 5.1400\ldots\) awrt (5.09 ~ 5.14) | A1 | ||||||||||||||||||||||||||||||
| \(\nu = 5 - 3 = 2\) (for 5 – 3 or 2 can be implied by 5.991 seen) | B1 | ||||||||||||||||||||||||||||||
| \(\chi^2_2(0.05) = 5.991\) | B1 | ||||||||||||||||||||||||||||||
| \(5.09 \lt 5.991\) so insufficient evidence to reject \(\mathrm{H}_0\) | M1 | ||||||||||||||||||||||||||||||
| Normal distribution is a good fit. | A1 | ||||||||||||||||||||||||||||||
| (8) | |||||||||||||||||||||||||||||||
| (14 marks) |
Notes
1st B1 for both hypotheses. B0 if they include 5.49 or 6.88. Condone \(X\)~\(\mathrm{N}(\mu, \sigma^2)\) etc
1st M1 for attempting \(\dfrac{(O - E)^2}{E}\) or \(\dfrac{O^2}{E}\), at least 3 correct expressions or values.
1st A1 for at least 4 correct calcs - 3rd or 4th column. (2 dp or better and allow e.g. 7.47)
Allow any value in the ranges for the last two rows.
2nd A1 for a test statistic that is awrt 5.09 ~ 5.14. Award M1A1A1 if this is obtained.
2nd M1 for a correct statement based on their test statistic ( > 1) and their cv (> 3.8)
Contradictory statements score M0 e.g. “significant” do not reject \(\mathrm{H}_0\).
3rd A1 for a correct comment suggesting that normal model is suitable or manager’s belief is correct. No f t . Condone mention of 5.49 or 6.88 here. Hypotheses wrong way round scores A0