Bottles of Lanta contain approximately $$\(300 \mathrm{ml}\)$$ of juice. The volume of juice, in millilitres, in a bottle is $$\(300+X\)$$, where $$\(X\)$$ is a random variable with probability density function given by $$\[ f(x)= \begin{cases}\frac{3}{4000}\left(100-x^{2}\right) & -10 \leqslant x \leqslant 10, \\ 0 & \text { otherwise. }\end{cases} \]$$ Find the probability that a randomly chosen bottle of Lanta contains more than $$\(305 \mathrm{ml}\)$$ of juice. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................
Exam No:9709_m20_qp_62 Year:2020 Question No:5(a)
Answer:
\(\frac{3}{4000} \int_{5}^{10}\left(100-x^{2}\right) \mathrm{d} x\)
\(=\frac{3}{4000}\left[100 x-\frac{x^{3}}{3}\right]_{5}^{10}\)
\(=\frac{3}{4000}\left(1000-\frac{1000}{3}-500+\frac{125}{3}\right)\)
\(=0.156(3 \mathrm{sf})\) or \(\frac{5}{32}\)
\(=\frac{3}{4000}\left[100 x-\frac{x^{3}}{3}\right]_{5}^{10}\)
\(=\frac{3}{4000}\left(1000-\frac{1000}{3}-500+\frac{125}{3}\right)\)
\(=0.156(3 \mathrm{sf})\) or \(\frac{5}{32}\)
Knowledge points:
6.3.2 use a probability density function to solve problems involving probabilities, and to calculate the mean and variance of a distribution. (Including location of the median or other percentiles of a distribution by direct consideration of an area using the density function.) (Explicit knowledge of the cumulative distribution function is not included.)
Solution:
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