Standard +0.3 This is a straightforward application of linear combinations of normal distributions requiring students to form 6C + 6S + 550, find its mean and variance using standard formulas, then calculate a single probability. While it involves multiple random variables, the method is routine for S2 students with no conceptual challenges beyond applying learned formulas correctly.
3 Weights of cups have a normal distribution with mean 91 g and standard deviation 3.2 g . Weights of saucers have an independent normal distribution with mean 72 g and standard deviation 2.6 g . Cups and saucers are chosen at random to be packed in boxes, with 6 cups and 6 saucers in each box. Given that each empty box weighs 550 g , find the probability that the total weight of a box containing 6 cups and 6 saucers exceeds 1550 g .
For mean (1528) oe and variance (102). May be implied by use of \(N(1528, 10.1^2)\)
\(\dfrac{1550 - 1528}{\sqrt{102}} = 2.178\)
M1
For standardising. No SD/Var mix
\(1 - \Phi(2.178) = 0.0147\) (3 sf)
M1, A1 [5]
For correct area consistent with working
## Question 3:
| Answer | Mark | Guidance |
|--------|------|----------|
| $\text{Var(total)} = 6(3.2^2 + 2.6^2)(+0) = 102$; Total $\sim N(1528, 102)$ | B1, B1 | For mean (1528) oe and variance (102). May be implied by use of $N(1528, 10.1^2)$ |
| $\dfrac{1550 - 1528}{\sqrt{102}} = 2.178$ | M1 | For standardising. No SD/Var mix |
| $1 - \Phi(2.178) = 0.0147$ (3 sf) | M1, A1 [5] | For correct area consistent with working |
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3 Weights of cups have a normal distribution with mean 91 g and standard deviation 3.2 g . Weights of saucers have an independent normal distribution with mean 72 g and standard deviation 2.6 g . Cups and saucers are chosen at random to be packed in boxes, with 6 cups and 6 saucers in each box. Given that each empty box weighs 550 g , find the probability that the total weight of a box containing 6 cups and 6 saucers exceeds 1550 g .
\hfill \mbox{\textit{CAIE S2 2013 Q3 [5]}}