On any given day, the probability that Moena messages her friend Pasha is $$\(0.72\)$$. Use an approximation to find the probability that in any period of 100 days Moena messages Pasha on fewer than 64 days. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................
Exam No：9709_s20_qp_52 Year：2020 Question No：7(c)
Answer:
$
\text { Mean }=100 \times 0.72=72
$
$
\operatorname{Var}=100 \times 0.72 \times 0.28=20.16
$
\(\mathrm{P}(\) less than 64 \()=\mathrm{P}\left(z<\frac{63.5-72}{\sqrt{20.16}}\right)\)
(M1 for substituting their \(\mu\) and \(\sigma\) into \(\pm\) standardisation formula with a numerical value for ' \(63.5\) ')
Using either \(63.5\) or \(64.5\) within \(\mathrm{a} \pm\) standardisation formula
Appropriate area \(\Phi\), from standardisation formula \(\mathrm{P}(z<\ldots)\) in final solution \(=\mathrm{P}(z<-1.893)\)
$
0.0292
$
\text { Mean }=100 \times 0.72=72
$
$
\operatorname{Var}=100 \times 0.72 \times 0.28=20.16
$
\(\mathrm{P}(\) less than 64 \()=\mathrm{P}\left(z<\frac{63.5-72}{\sqrt{20.16}}\right)\)
(M1 for substituting their \(\mu\) and \(\sigma\) into \(\pm\) standardisation formula with a numerical value for ' \(63.5\) ')
Using either \(63.5\) or \(64.5\) within \(\mathrm{a} \pm\) standardisation formula
Appropriate area \(\Phi\), from standardisation formula \(\mathrm{P}(z<\ldots)\) in final solution \(=\mathrm{P}(z<-1.893)\)
$
0.0292
$
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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