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SPM Additional Mathematics 2023, Paper 1 (Question 12)


Question 12:
A farm produce strawberries. The masses of the strawberries, x grams, are normally distributed such that X∼ N[20, (0.5)2].

(a) State the mean and the variance of the distribution. [1 mark]

(b) Diagram 8 shows the standard normal distribution graph for mass of the strawberries. The shaded region represents the probability that the mass of the strawberries is between (-α + 40) grams and α grams.


(i) If the shaded region represents P( |Z∣ < k), express m in terms of α.

(ii) Find the value of α when P( |Z∣ < k) = 0.2662.
[3 marks]

Solution:
(a)
$$ \begin{aligned} & X \sim N\left(\mu, \sigma^2\right) \\ & X \sim N\left(20,0.5^2\right) \\ & \text { Mean }=20 \\ & \text { Variance }=(0.5)^2=0.25 \end{aligned} $$

(b)(i)
$$ \begin{aligned} &\begin{aligned} X & \sim N\left(20,0.5^2\right) \\ Z & =\frac{x-\mu}{\sigma} \end{aligned}\\ &\text { Given that the shaded region }\\ &\begin{aligned} & =P(-\alpha+40<X<\alpha) \\ & =P\left(\frac{-\alpha+40-20}{0.5}<Z<\frac{\alpha-20}{0.5}\right) \\ & =P\left(\frac{-\alpha+20}{0.5}<Z<\frac{\alpha-20}{0.5}\right) \end{aligned} \end{aligned} $$

Given also that the shaded region represents P(| Z ∣ < k).
$$ \begin{aligned} &P(-k<Z<k)=P(m<Z<k)\\ &\text { Compare with shaded region above, }\\ &\begin{aligned} & m=\frac{-\alpha+20}{0.5} \\ & m=\frac{(-\alpha+20) \times 2}{0.5 \times 2} \\ & m=40-2 \alpha \end{aligned} \end{aligned} $$

(b)(ii)
$$ \begin{aligned} P(|Z|<k) & =0.2662 \\ P(-k<Z<k) & =0.2662 \\ 1-P(Z<-k)-P(Z>k) & =0.2662 \\ 1-P(Z>k)-P(Z>k) & =0.2662 \\ 1-2 P(Z>k) & =0.2662 \\ 2 P(Z>k) & =0.7338 \\ P(Z>k) & =0.3669 \end{aligned} $$


$$ \begin{aligned} &\text { From the table, } k=0.34\\ &\begin{aligned} m & =40-2 \alpha[\text { From } b(i i)] \\ -0.34 & =40-2 \alpha \\ 2 \alpha & =40.34 \\ \alpha & =20.17 \end{aligned} \end{aligned} $$

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