Customers arrive at a particular shop at random times. It has been found that the mean number of customers who arrive during a 5-minute interval is $$\(2.1\)$$. Use a suitable approximating distribution to find the probability that fewer than 40 customers arrive during a 2-hour interval. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................
Exam No：9709_s21_qp_62 Year：2021 Question No：7(c)
Answer:
\(\mathrm{N}(50.4,50.4)\)
\(\frac{39.5-50.4}{\sqrt{50.4}}[=-1.535]\)
\(\Phi\left({ }^{\prime}-1.535^{\prime}\right)=1-\Phi\left({ }^{\prime} 1.535^{\prime}\right)\)
\(0.0624(3\) sf) or \(0.0623\)
\(\frac{39.5-50.4}{\sqrt{50.4}}[=-1.535]\)
\(\Phi\left({ }^{\prime}-1.535^{\prime}\right)=1-\Phi\left({ }^{\prime} 1.535^{\prime}\right)\)
\(0.0624(3\) sf) or \(0.0623\)
Knowledge points:
6.1.5 use the normal distribution, with continuity correction, as an approximation to the Poisson distribution where appropriate. (The condition that is large should be known; , approximately.)
Solution:
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