IB Mathematics AHL 4.17 Poisson distribution -AI HL Paper 3- Exam Style Questions- New Syllabus
Question
This question employs statistical tests to assess whether advertising enhances profits for a grocery store.
Suwannee, the manager of a grocery store in Nong Khai, conducts a statistical analysis on the daily sales of bags of rice. She gathers sample data by recording the number of bags of rice sold daily over 90 days.
| Number of bags of rice sold | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| Number of days | 1 | 8 | 12 | 11 | 19 | 14 | 13 | 8 | 2 | 0 | 2 |
She believes the data follows a Poisson distribution.
(a)(i) Calculate the mean and variance of the sample data provided in the table. [2]
(a)(ii) Hence, explain why Suwannee believes the data follows a Poisson distribution. [1]
(b) Identify one assumption Suwannee must make about rice sales to support her belief in a Poisson distribution. [1]
Suwannee knows from historical records that the store sells an average of 4.2 bags of rice daily. The following table presents the expected frequencies of bags of rice sold daily over the 90-day period, assuming a Poisson distribution with mean 4.2.
| Number of bags of rice sold | \(\leq 1\) | 2 | 3 | 4 | 5 | 6 | 7 | \(\geq 8\) |
|---|---|---|---|---|---|---|---|---|
| Expected frequency | \( a \) | 11.903 | 16.665 | \( b \) | 14.698 | 10.289 | 6.173 | \( c \) |
(c) Calculate the values of \( a \), \( b \), and \( c \), rounded to three decimal places. [5]
Suwannee conducts a \(\chi^2\) goodness of fit test at the 5% significance level to determine if the data follows a Poisson distribution with mean 4.2.
(d)(i) State the number of degrees of freedom for the test. [1]
(d)(ii) Conduct the \(\chi^2\) goodness of fit test and provide a conclusion with reasoning. [7]
Suwannee asserts that advertising in a local newspaper for 300 Thai Baht (THB) per day will increase rice sales, while Krit, the store owner, claims it will not increase overall profit.
Krit agrees to advertise for the next 60 days. During this period, Suwannee records that the store sells 282 bags of rice with a profit of 495 THB per bag.
(e)(i) Suwannee aims to perform a hypothesis test to determine if the number of bags of rice sold during the 60 days increased compared to historical records. Conduct this test at the 5% significance level by finding a critical value. [6]
(e)(ii) State the probability of a Type I error for this test. [1]
(f) Considering the claims of Suwannee and Krit, evaluate whether the advertising was beneficial to the store. [3]
▶️ Answer/Explanation
Markscheme
(a)(i)
Observed frequencies: \( (1, 8, 12, 11, 19, 14, 13, 8, 2, 0, 2) \), sum = 90.
Mean:
\( \bar{x} = \frac{0(1) + 1(8) + 2(12) + 3(11) + 4(19) + 5(14) + 6(13) + 7(8) + 8(2) + 9(0) + 10(2)}{90} = \frac{381}{90} \approx 4.233 \).
Variance:
\( s^2 = \frac{0^2(1) + 1^2(8) + 2^2(12) + 3^2(11) + 4^2(19) + 5^2(14) + 6^2(13) + 7^2(8) + 8^2(2) + 9^2(0) + 10^2(2)}{90} – (4.233)^2 \approx \frac{1632}{90} – 17.922 \approx 4.268 \).
Mean \( \approx 4.23 \), Variance \( \approx 4.27 \) A1A1
[2 marks]
Observed frequencies: \( (1, 8, 12, 11, 19, 14, 13, 8, 2, 0, 2) \), sum = 90.
Mean:
\( \bar{x} = \frac{0(1) + 1(8) + 2(12) + 3(11) + 4(19) + 5(14) + 6(13) + 7(8) + 8(2) + 9(0) + 10(2)}{90} = \frac{381}{90} \approx 4.233 \).
Variance:
\( s^2 = \frac{0^2(1) + 1^2(8) + 2^2(12) + 3^2(11) + 4^2(19) + 5^2(14) + 6^2(13) + 7^2(8) + 8^2(2) + 9^2(0) + 10^2(2)}{90} – (4.233)^2 \approx \frac{1632}{90} – 17.922 \approx 4.268 \).
Mean \( \approx 4.23 \), Variance \( \approx 4.27 \) A1A1
[2 marks]
(a)(ii)
For a Poisson distribution, the mean equals the variance. Since \( \bar{x} \approx 4.23 \) and \( s^2 \approx 4.27 \) are close, the data supports a Poisson distribution.
Mean and variance are approximately equal A1
[1 mark]
For a Poisson distribution, the mean equals the variance. Since \( \bar{x} \approx 4.23 \) and \( s^2 \approx 4.27 \) are close, the data supports a Poisson distribution.
Mean and variance are approximately equal A1
[1 mark]
(b)
The number of bags sold each day is independent, and sales occur at a constant average rate.
Independence and constant rate A1
[1 mark]
The number of bags sold each day is independent, and sales occur at a constant average rate.
Independence and constant rate A1
[1 mark]
(c)
For \( X \sim \text{Poisson}(4.2) \), probabilities: \( \mathbb{P}(X=k) = e^{-4.2} \frac{4.2^k}{k!} \).
\( a = 90 \left( \mathbb{P}(X=0) + \mathbb{P}(X=1) \right) = 90 \left( e^{-4.2} + 4.2 e^{-4.2} \right) \approx 90 (0.014996 + 0.062981) \approx 7.018 \).
\( b = 90 \mathbb{P}(X=4) = 90 \left( e^{-4.2} \frac{4.2^4}{4!} \right) \approx 90 (0.19442) \approx 17.498 \).
\( c = 90 \left( 1 – \sum_{k=0}^{7} \mathbb{P}(X=k) \right) \approx 90 (0.06394) \approx 5.755 \).
\( a \approx 7.018 \), \( b \approx 17.498 \), \( c \approx 5.755 \) M1A1A1A1A1
[5 marks]
For \( X \sim \text{Poisson}(4.2) \), probabilities: \( \mathbb{P}(X=k) = e^{-4.2} \frac{4.2^k}{k!} \).
\( a = 90 \left( \mathbb{P}(X=0) + \mathbb{P}(X=1) \right) = 90 \left( e^{-4.2} + 4.2 e^{-4.2} \right) \approx 90 (0.014996 + 0.062981) \approx 7.018 \).
\( b = 90 \mathbb{P}(X=4) = 90 \left( e^{-4.2} \frac{4.2^4}{4!} \right) \approx 90 (0.19442) \approx 17.498 \).
\( c = 90 \left( 1 – \sum_{k=0}^{7} \mathbb{P}(X=k) \right) \approx 90 (0.06394) \approx 5.755 \).
\( a \approx 7.018 \), \( b \approx 17.498 \), \( c \approx 5.755 \) M1A1A1A1A1
[5 marks]
(d)(i)
With 8 categories (\( \leq 1, 2, 3, 4, 5, 6, 7, \geq 8 \)) and no parameters estimated, degrees of freedom = \( 8 – 1 = 7 \).
Degrees of freedom: 7 A1
[1 mark]
With 8 categories (\( \leq 1, 2, 3, 4, 5, 6, 7, \geq 8 \)) and no parameters estimated, degrees of freedom = \( 8 – 1 = 7 \).
Degrees of freedom: 7 A1
[1 mark]
(d)(ii)
Observed: \( (9, 12, 11, 19, 14, 13, 8, 4) \).
Expected: \( (7.018, 11.903, 16.665, 17.498, 14.698, 10.289, 6.173, 5.755) \).
Test statistic:
\( \chi^2 = \sum \frac{(O – E)^2}{E} = \frac{(9-7.018)^2}{7.018} + \frac{(12-11.903)^2}{11.903} + \cdots + \frac{(4-5.755)^2}{5.755} \approx 4.44 \).
For \( df = 7 \), \( p \)-value \( \approx 0.728 \).
Since \( p > 0.05 \) (or \( \chi^2 = 4.44 < \chi^2_{0.95,7} = 14.07 \)), do not reject \( H_0 \).
Conclusion: The data is consistent with a Poisson distribution with mean 4.2 M1A1A1A1A1A1A1
[7 marks]
Observed: \( (9, 12, 11, 19, 14, 13, 8, 4) \).
Expected: \( (7.018, 11.903, 16.665, 17.498, 14.698, 10.289, 6.173, 5.755) \).
Test statistic:
\( \chi^2 = \sum \frac{(O – E)^2}{E} = \frac{(9-7.018)^2}{7.018} + \frac{(12-11.903)^2}{11.903} + \cdots + \frac{(4-5.755)^2}{5.755} \approx 4.44 \).
For \( df = 7 \), \( p \)-value \( \approx 0.728 \).
Since \( p > 0.05 \) (or \( \chi^2 = 4.44 < \chi^2_{0.95,7} = 14.07 \)), do not reject \( H_0 \).
Conclusion: The data is consistent with a Poisson distribution with mean 4.2 M1A1A1A1A1A1A1
[7 marks]
(e)(i)
[6 marks]
Method 1 (Exact Poisson Test):
Historical mean: 4.2 bags/day. Over 60 days: \( \mu = 60 \times 4.2 = 252 \).
Hypotheses: \( H_0: \mu = 252 \), \( H_1: \mu > 252 \) (one-tailed).
Observed: \( x = 282 \).
Find smallest \( c \) where \( P(X \geq c \mid \mu = 252) \leq 0.05 \):
\( P(X \geq 277) \approx 0.06305 > 0.05 \),
\( P(X \geq 278) \approx 0.05584 > 0.05 \),
\( P(X \geq 279) \approx 0.04931 \leq 0.05 \).
Critical value: \( c = 279 \).
Since \( 282 \geq 279 \), reject \( H_0 \).
Historical mean: 4.2 bags/day. Over 60 days: \( \mu = 60 \times 4.2 = 252 \).
Hypotheses: \( H_0: \mu = 252 \), \( H_1: \mu > 252 \) (one-tailed).
Observed: \( x = 282 \).
Find smallest \( c \) where \( P(X \geq c \mid \mu = 252) \leq 0.05 \):
\( P(X \geq 277) \approx 0.06305 > 0.05 \),
\( P(X \geq 278) \approx 0.05584 > 0.05 \),
\( P(X \geq 279) \approx 0.04931 \leq 0.05 \).
Critical value: \( c = 279 \).
Since \( 282 \geq 279 \), reject \( H_0 \).
Method 2 (Normal Approximation):
Sample mean: \( \bar{x} = \frac{282}{60} = 4.7 \).
Hypotheses: \( H_0: \mu = 4.2 \), \( H_1: \mu > 4.2 \).
For Poisson(\( \mu = 4.2 \)), variance = 4.2. Standard error: \( \sqrt{\frac{4.2}{60}} \approx \sqrt{0.07} \approx 0.2646 \).
Critical value: \( \mu + z_{0.05} \cdot \text{SE} = 4.2 + 1.645 \times 0.2646 \approx 4.2 + 0.43518 \approx 4.635 \).
Since \( 4.7 > 4.635 \), reject \( H_0 \).
Conclusion: There is evidence at the 5% level that advertising increased sales M1A1A1A1A1A1 Sample mean: \( \bar{x} = \frac{282}{60} = 4.7 \).
Hypotheses: \( H_0: \mu = 4.2 \), \( H_1: \mu > 4.2 \).
For Poisson(\( \mu = 4.2 \)), variance = 4.2. Standard error: \( \sqrt{\frac{4.2}{60}} \approx \sqrt{0.07} \approx 0.2646 \).
Critical value: \( \mu + z_{0.05} \cdot \text{SE} = 4.2 + 1.645 \times 0.2646 \approx 4.2 + 0.43518 \approx 4.635 \).
Since \( 4.7 > 4.635 \), reject \( H_0 \).
[6 marks]
(e)(ii)
Type I error probability (rejecting \( H_0 \) when \( \mu = 252 \)) is the test size:
\( P(X \geq 279 \mid \mu = 252) \approx 0.0493 \).
Probability \( \approx 0.0493 \) A1
[1 mark]
Type I error probability (rejecting \( H_0 \) when \( \mu = 252 \)) is the test size:
\( P(X \geq 279 \mid \mu = 252) \approx 0.0493 \).
Probability \( \approx 0.0493 \) A1
[1 mark]
(f)
Profit per bag = 495 THB. Advertising cost = \( 300 \times 60 = 18,000 \) THB.
Without advertising: Expected sales = \( 60 \times 4.2 = 252 \) bags, profit = \( 252 \times 495 = 124,740 \) THB.
With advertising: Sales = 282 bags, profit = \( 282 \times 495 = 139,590 \) THB.
Extra bags = \( 282 – 252 = 30 \). Extra profit = \( 30 \times 495 = 14,850 \) THB.
Net change = \( 14,850 – 18,000 = -3,150 \) THB.
Suwannee’s claim (increased sales) is supported, but Krit’s claim (no profit increase) holds due to the net loss.
Advertising was not beneficial due to a net loss of 3,150 THB M1A1A1
[3 marks]
Profit per bag = 495 THB. Advertising cost = \( 300 \times 60 = 18,000 \) THB.
Without advertising: Expected sales = \( 60 \times 4.2 = 252 \) bags, profit = \( 252 \times 495 = 124,740 \) THB.
With advertising: Sales = 282 bags, profit = \( 282 \times 495 = 139,590 \) THB.
Extra bags = \( 282 – 252 = 30 \). Extra profit = \( 30 \times 495 = 14,850 \) THB.
Net change = \( 14,850 – 18,000 = -3,150 \) THB.
Suwannee’s claim (increased sales) is supported, but Krit’s claim (no profit increase) holds due to the net loss.
Advertising was not beneficial due to a net loss of 3,150 THB M1A1A1
[3 marks]