2 years ago

What is the volume of a sphere with a diameter of 23.5 in, rounded to the nearest

Mathematics

1 answer:

zlopas [31]2 years ago

5 0

Answer:

V≈6795.2in³

Step-by-step explanation:

I think

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For X a binomial random variable with n=5 and p=1/4, answer the following

mash [69]

Recall that for $X\sim\mathrm{Bin}(n,p)$ , i.e. a random variable $X$ following a binomial distribution over $n$ trials and with probability parameter $p$ ,

$\mathbb P(X=x)=f_X(x)=\begin{cases}\dbinom nx p^x(1-p)^{n-x}&\text{for }x\in\{0,1,\ldots,n\}\\\\0&\text{otherwise}\end{cases}$

So you have

$\mathbb P(X=2)=\dbinom52\left(\dfrac14\right)^2\left(1-\dfrac34\right)^{5-2}=\dfrac{135}{512}\approx0.26$

$\mathbb P(2\le X\le4)=\displaystyle\sum_{x=2}^4\mathbb P(X=x)$
$=\dbinom52\left(\dfrac14\right)^2\left(1-\dfrac14\right)^{5-2}+\dbinom52\left(\dfrac14\right)^3\left(1-\dfrac14\right)^{5-3}+\dbinom52\left(\dfrac14\right)^4\left(1-\dfrac14\right)^{5-4}$
$=\dfrac{375}{1024}\approx0.37$

The expected value of $X\sim\mathrm{Bin}(n,p)$ is simply $np$ , while the standard deviation is $\sqrt{np(1-p)}$ . In this case, they are $\dfrac54=1.25$ and $\sqrt{\dfrac{15}{16}}\approx0.97$ , respectively.