Question

suppose a random variable, x, arises from a binomial experiment. suppose n = 7, and p = 0.53. write the probability distribution. round to six decimal places, if necessary. select the correct histogram.

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 on a single - trial, and $C(n,k)=\frac{n!}{k!(n - k)!}$. Here, $n = 7$ and $p=0.53$, so $1 - p = 0.47$.

Step2: Calculate $P(X = 0)$

$C(7,0)=\frac{7!}{0!(7 - 0)!}=1$, $P(X = 0)=C(7,0)\times(0.53)^{0}\times(0.47)^{7}=1\times1\times0.47^{7}\approx0.005654$.

Step3: Calculate $P(X = 1)$

$C(7,1)=\frac{7!}{1!(7 - 1)!}=\frac{7!}{1!6!}=7$, $P(X = 1)=C(7,1)\times(0.53)^{1}\times(0.47)^{6}=7\times0.53\times0.47^{6}\approx0.042597$.

Step4: Calculate $P(X = 2)$

$C(7,2)=\frac{7!}{2!(7 - 2)!}=\frac{7\times6}{2\times1}=21$, $P(X = 2)=C(7,2)\times(0.53)^{2}\times(0.47)^{5}=21\times0.53^{2}\times0.47^{5}\approx0.137637$.

Step5: Calculate $P(X = 3)$

$C(7,3)=\frac{7!}{3!(7 - 3)!}=\frac{7\times6\times5}{3\times2\times1}=35$, $P(X = 3)=C(7,3)\times(0.53)^{3}\times(0.47)^{4}=35\times0.53^{3}\times0.47^{4}\approx0.253570$.

Step6: Calculate $P(X = 4)$

$C(7,4)=\frac{7!}{4!(7 - 4)!}=35$, $P(X = 4)=C(7,4)\times(0.53)^{4}\times(0.47)^{3}=35\times0.53^{4}\times0.47^{3}\approx0.290714$.

Step7: Calculate $P(X = 5)$

$C(7,5)=\frac{7!}{5!(7 - 5)!}=21$, $P(X = 5)=C(7,5)\times(0.53)^{5}\times(0.47)^{2}=21\times0.53^{5}\times0.47^{2}\approx0.206147$.

Step8: Calculate $P(X = 6)$

$C(7,6)=\frac{7!}{6!(7 - 6)!}=7$, $P(X = 6)=C(7,6)\times(0.53)^{6}\times(0.47)^{1}=7\times0.53^{6}\times0.47\approx0.087393$.

Step9: Calculate $P(X = 7)$

$C(7,7)=\frac{7!}{7!(7 - 7)!}=1$, $P(X = 7)=C(7,7)\times(0.53)^{7}\times(0.47)^{0}=1\times0.53^{7}\times1\approx0.015007$.

Answer:

0: 0.005654
1: 0.042597
2: 0.137637
3: 0.253570
4: 0.290714
5: 0.206147
6: 0.087393
7: 0.015007