There are 400 students at a school in a certain country. Each student was asked whether they preferred swimming, cycling or running and the results are given in the following table. A student is chosen at random. On average at all the schools in this country 30% of the students do not like any sports. 90 students from this country are now chosen at random. Use an approximation to find the probability that fewer than 32 of them do not like any sports. ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................
Exam No：9709_m21_qp_52 Year：2021 Question No：7(b)(ii)
Answer:
\([\mathrm{p}=0 \cdot 3]\)
Mean \(=0.3 \times 90=27\);
variance \(=0.3 \times 90 \times 0.7=18.9\)
\(\mathrm{P}(X<32)=P\left(z<\frac{31.5-27}{\sqrt{18.9}}\right)\)
\(=\Phi(1.035)\)
\(=0 \cdot 850\)
Mean \(=0.3 \times 90=27\);
variance \(=0.3 \times 90 \times 0.7=18.9\)
\(\mathrm{P}(X<32)=P\left(z<\frac{31.5-27}{\sqrt{18.9}}\right)\)
\(=\Phi(1.035)\)
\(=0 \cdot 850\)
Knowledge points:
5.5.3 recall conditions under which the normal distribution can be used as an approximation to the binomial distribution, and use this approximation, with a continuity correction, in solving problems. (n sufficiently large to ensure that both np > 5 and nq > 5.)
Solution:
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