Answer to Question #103299 in Statistics and Probability for Maow Abdullahi

Question

b).For the past records, about 40% of a firms orders are for export . Their record for export is 48% in one particular financial quarter. If they expect to sastify about 80 orders in the next financial quarter, what is the probability that they will break their previous export record?.

Answer 1

For large values of n, the distributions of the count X and the sample proportion $\hat{p}$ are approximately normal. In practice, the approximation is adequate provided that both $np\geq5$ and $n(1-p)\geq10$ , since there is then enough symmetry in the underlying binomial distribution.

Given the proportion $p$ is expected to be 0.48 and from past records $\hat{p}$ was found to be 0.4 and $n=10.$ Then

$np=10(0.48)=4.8<5,$

$n(1-p)=10(1-0.48)=5.2<10$

Normal distribution cannot be used.

Binomial distribution is appropriate for this event. Then $10(0.4)=4$

Probability of breaking the record

$P(X>4)=1-P(X=0)-P(X=1)-$ $P(X=2)-P(X=3)-P(X=4)=$

$=1-\dbinom{10}{0}(0.48)^0(1-0.48)^{10-0}-\dbinom{10}{1}(0.48)^1(1-0.48)^{10-1}-$

$-\dbinom{10}{2}(0.48)^2(1-0.48)^{10-2}-\dbinom{10}{3}(0.48)^3(1-0.48)^{10-3}-$

$-\dbinom{10}{4}(0.48)^4(1-0.48)^{10-4}\approx$

$\approx1-0.0015-0.0133-0.0554-0.1364-0.2204\approx$

$\approx0.573$