Answer to Question #203062 in Statistics and Probability for Erick

Question

A survey claims that 73% of Fiji households have at least one credit card. Find the probability

that in a random sample of 550 Fiji households, 375 to 385 have at least one credit card.

Answer 1

Let $X=$ the number of households having at least one credit card: $X\sim Bin(n, p)$

$P(X=x)=\dbinom{n}{x}p^x(1-p)^{n-x}$

Given $p=0.73, n=550$

$P(X=375)=\dbinom{550}{375}0.73^{375}(1-0.73)^{550-375}$

$\approx 0.00159641506$

$P(X=376)=\dbinom{550}{376}0.73^{376}(1-0.73)^{550-376}$

$\approx 0.00200888518$

$P(X=377)=\dbinom{550}{377}0.73^{377}(1-0.73)^{550-377}$

$\approx 0.00250681398$

$P(X=378)=\dbinom{550}{378}0.73^{378}(1-0.73)^{550-378}$

$\approx0.0031019551$

$P(X=379)=\dbinom{550}{379}0.73^{379}(1-0.73)^{550-379}$

$\approx 0.00380613195$

$P(X=380)=\dbinom{550}{380}0.73^{380}(1-0.73)^{550-380}$

$\approx 0.00463079388$

$P(X=381)=\dbinom{550}{381}0.73^{381}(1-0.73)^{550-381}$

$\approx 0.00558648314$

$P(X=382)=\dbinom{550}{382}0.73^{382}(1-0.73)^{550-382}$

$\approx 0.00668222246$

$P(X=383)=\dbinom{550}{383}0.73^{383}(1-0.73)^{550-383}$

$\approx 0.00792484056$

$P(X=384)=\dbinom{550}{384}0.73^{384}(1-0.73)^{550-384}$

$\approx 0.00931826112$

$P(X=385)=\dbinom{550}{385}0.73^{385}(1-0.73)^{550-385}$

$\approx 0.01086278867$

$P(375\leq X\leq 385)\approx0.0580255911$