IB Mathematics AHL 4.14 Linear transformation AI HL Paper 3- Exam Style Questions- New Syllabus

(a)
First, find the probability that a single egg weighs at least 55.0 g:
\( X \sim N(52.0, 3.7^2) \)
\( P(X \geq 55.0) = P\left(Z \geq \frac{55.0 – 52.0}{3.7}\right) = P(Z \geq 0.8108…) \)
Using normal distribution tables or GDC: \( P(Z \geq 0.8108) = 0.208737… \)

Let \( Y \) be the number of eggs (out of 25) that weigh at least 55.0 g.
Then \( Y \sim B(25, 0.208737…) \)
We want \( P(Y \geq 5) = 1 – P(Y \leq 4) \)
Using binomial cumulative distribution:
\( P(Y \geq 5) = 0.621150… \approx 0.621 \)
\( \boxed{0.621} \)

(b)
Let \( \bar{X} \) be the sample mean weight of 25 eggs.
By Central Limit Theorem: \( \bar{X} \sim N\left(52.0, \frac{3.7^2}{25}\right) \)
So variance = \( \frac{3.7^2}{25} \) and standard deviation = \( \frac{3.7}{5} = 0.74 \)

We want \( P(51 < \bar{X} < 53) \):
Standardize: \( z_1 = \frac{51 – 52.0}{0.74} = -1.3514 \)
\( z_2 = \frac{53 – 52.0}{0.74} = 1.3514 \)
\( P(51 < \bar{X} < 53) = P(-1.3514 < Z < 1.3514) \)
Using normal distribution: \( = 0.823417… \approx 0.823 \)
\( \boxed{0.823} \)

(c)
Step 1: State hypotheses
\( H_0: \mu = 52.0 \) (population mean weight is 52.0 g)
\( H_1: \mu > 52.0 \) (population mean weight has increased)

Step 2: Calculate sample statistics from Table 1
Sample size \( n = 10 \)
Sample mean \( \bar{x} = \frac{54.5 + 52.4 + 50.4 + 54.3 + 54.2 + 53.1 + 50.2 + 51.8 + 52.2 + 53.5}{10} = 52.66 \)
Sample standard deviation \( s = 1.668… \) (using GDC)

Step 3: Perform one-sample t-test
Test statistic: \( t = \frac{\bar{x} – \mu_0}{s/\sqrt{n}} = \frac{52.66 – 52.0}{1.668/\sqrt{10}} = 1.250… \)
Degrees of freedom = \( n – 1 = 9 \)
p-value = \( P(T_9 > 1.250…) = 0.105055… \)

Step 4: Compare p-value with significance level
\( 0.105 > 0.05 \)
Since p-value > 0.05, we fail to reject \( H_0 \)

Step 5: Conclusion
There is insufficient evidence at the 5% significance level to conclude that the mean weight of eggs has increased from 52.0 g.
\( \boxed{\text{Insufficient evidence to reject } H_0} \)

(d)
For 99% confidence interval with \( n = 10 \), \( \bar{x} = 52.66 \), \( s = 1.668 \):
Degrees of freedom = 9, \( t^*_{0.005, 9} = 3.2498 \) (from t-distribution table)

Confidence interval: \( \bar{x} \pm t^* \times \frac{s}{\sqrt{n}} \)
\( = 52.66 \pm 3.2498 \times \frac{1.668}{\sqrt{10}} \)
\( = 52.66 \pm 1.713… \)
\( = (50.95, 54.37) \)

Or using GDC directly: (51.0712…, 54.2488…)
\( \boxed{(51.1, 54.2)} \) (to 3 significant figures)

(e)
(i) Since the same 8 chickens are measured before and after, this is a paired t-test.
\( \boxed{\text{Paired t-test}} \)

(ii) One necessary assumption: The differences in weights are normally distributed.
\( \boxed{\text{Differences are normally distributed}} \)

(f)
Step 1: Calculate differences (after – before)
Differences: 1.3, 0.4, 0.7, 2.5, 0.9, -1.5, 1.0, 2.0
Mean difference \( \bar{d} = 0.9125 \)
Standard deviation of differences \( s_d = 1.093… \)

Step 2: State hypotheses
\( H_0: \mu_d = 0 \) (no change in mean weight)
\( H_1: \mu_d > 0 \) (mean weight has increased)

Step 3: Perform paired t-test
Test statistic: \( t = \frac{\bar{d} – 0}{s_d/\sqrt{n}} = \frac{0.9125}{1.093/\sqrt{8}} = 2.360… \)
Degrees of freedom = 7
p-value = \( P(T_7 > 2.360…) = 0.0250… \)

Step 4: Compare p-value with significance level
\( 0.0250 < 0.05 \)
Since p-value < 0.05, we reject \( H_0 \)

Step 5: Conclusion in context
There is sufficient evidence at the 5% significance level to conclude that the mean weight of the chickens’ eggs has increased following the new feed.
\( \boxed{\text{Reject } H_0 \text{, mean weight has increased}} \)

(g)
From Table 1 data, calculate unbiased variance estimate:
Using GDC or formula: \( s_G^2 = \frac{\sum (x_i – \bar{x})^2}{n-1} \)
With \( n = 10 \), \( \bar{x} = 52.66 \):
\( s_G^2 = 2.391555… \)
\( \boxed{2.39} \) (to 3 significant figures)

(h)
For F-test: \( F = \frac{s_R^2}{s_G^2} \)
Critical value \( F_{\text{crit}} = 3.1789 \) at 5% level
For t-test to be valid (equal variances), we need \( F \leq F_{\text{crit}} \)
So \( \frac{s_R^2}{s_G^2} \leq 3.1789 \)
\( s_R^2 \leq 3.1789 \times s_G^2 \)
\( s_R^2 \leq 3.1789 \times 2.391555… \)
\( s_R^2 \leq 7.602… \)

Maximum value of \( s_R^2 \) = 7.60 (to 3 significant figures)
\( \boxed{7.60} \)