Question

chowdary valluru 21 minutes ago bell ringer - sep 30 question 1) the random variable x has binomial distribution b(20, 0.43) determine each of the following: a) p(x = 6) b) p(x<9) c) p(x>=12) d) p(8<x<13) question 2) the probability of a customer ordering the silver color car is 0.2 find the probability that in the next 30 random orders a) exactly 10 orders are silver b) at most 8 orders are silver c) no more than 11 orders are silver

Explanation:

Step1: Recall binomial probability formula

The binomial probability formula is $P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}$, where $n$ is the number of trials, $k$ is the number of successes, $p$ is the probability of success in a single - trial, and $C(n,k)=\frac{n!}{k!(n - k)!}$.

Question 1

a) $P(X = 6)$ for $n = 20$ and $p=0.43$

Step1: Calculate combination $C(20,6)$

$C(20,6)=\frac{20!}{6!(20 - 6)!}=\frac{20!}{6!14!}=\frac{20\times19\times18\times17\times16\times15}{6\times5\times4\times3\times2\times1}=38760$

Step2: Calculate $P(X = 6)$

$P(X = 6)=C(20,6)\times(0.43)^{6}\times(1 - 0.43)^{20 - 6}$
$P(X = 6)=38760\times(0.43)^{6}\times(0.57)^{14}$
$P(X = 6)\approx38760\times0.006274224\times0.001487957\approx0.036$

b) $P(X\lt9)$

$P(X\lt9)=\sum_{k = 0}^{8}C(20,k)\times(0.43)^{k}\times(0.57)^{20 - k}$
We can use a binomial probability table or a calculator with binomial cumulative - distribution function. Using a calculator, $P(X\lt9)\approx0.496$

c) $P(X\geq12)$

$P(X\geq12)=1 - P(X\lt12)=1-\sum_{k = 0}^{11}C(20,k)\times(0.43)^{k}\times(0.57)^{20 - k}$
Using a calculator, $P(X\geq12)\approx0.077$

d) $P(8\lt X\lt13)$

$P(8\lt X\lt13)=P(X = 9)+P(X = 10)+P(X = 11)$
$P(X = 9)=C(20,9)\times(0.43)^{9}\times(0.57)^{11}$
$C(20,9)=\frac{20!}{9!11!}=167960$
$P(X = 9)\approx167960\times(0.43)^{9}\times(0.57)^{11}\approx0.155$
$P(X = 10)=C(20,10)\times(0.43)^{10}\times(0.57)^{10}$
$C(20,10)=184756$
$P(X = 10)\approx184756\times(0.43)^{10}\times(0.57)^{10}\approx0.179$
$P(X = 11)=C(20,11)\times(0.43)^{11}\times(0.57)^{9}$
$C(20,11)=167960$
$P(X = 11)\approx167960\times(0.43)^{11}\times(0.57)^{9}\approx0.157$
$P(8\lt X\lt13)\approx0.155 + 0.179+0.157 = 0.491$

Question 2

Here $n = 30$ and $p = 0.2$

a) Exactly 10 orders are silver

$C(30,10)=\frac{30!}{10!(30 - 10)!}=\frac{30!}{10!20!}=30045015$
$P(X = 10)=C(30,10)\times(0.2)^{10}\times(0.8)^{20}$
$P(X = 10)\approx30045015\times1.024\times10^{- 7}\times0.011529215\approx0.035$

b) At most 8 orders are silver

$P(X\leq8)=\sum_{k = 0}^{8}C(30,k)\times(0.2)^{k}\times(0.8)^{30 - k}$
Using a calculator, $P(X\leq8)\approx0.892$

c) No more than 11 orders are silver

$P(X\leq11)=\sum_{k = 0}^{11}C(30,k)\times(0.2)^{k}\times(0.8)^{30 - k}$
Using a calculator, $P(X\leq11)\approx0.994$

Answer:

Question 1:
a) $0.036$
b) $0.496$
c) $0.077$
d) $0.491$
Question 2:
a) $0.035$
b) $0.892$
c) $0.994$