How do I Find P ( 2 < X < 7 )? (Stats)

Question

How do I Find P ( 2 < X < 7 )? (Stats)

I Got the part A, but don't get how to do part B.

1 Answer

Impact of this question

2410 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-16T00:53:46

$P (2 < X < 7) = 54.43 %$ $P(2 < X < 7) = 54.43%$ .

Explanation:

The picture is a bit blurry, but it appears as though $X$ $X$ is the number of calls over a 1.5 minute interval. Thus, $X ~ Poisson (λ)$ $X" ~ Poisson"(lambda)$ where $λ = 1.5 \times 4 = 6 .$ $lambda = 1.5 xx 4 = 6.$

Thus, $P (X = x) = \frac{e^{- 6} \times 6^{x}}{x!}$ $P(X=x)" "=(e^-6 xx6^x)/(x!)$

Since $X$ $X$ is discrete, we know

$P (2 < X < 7) = 6 \sum x = 3 P (X = x)$ $P(2 < X < 7) = sum_(x=3)^6 P(X = x)$

$= P (X =3) + P (X =4) + P (X =5) + P (X =6)$ $= P(X"=3")+P(X"=4")+P(X"=5")+P(X"=6")$

We just need to find these 4 discrete probabilities.

$P (X = 3) = \frac{e^{- 6} \times 6^{3}}{3!} = 0.0892$ $P(X=3) = (e^-6 xx6^3)/(3!) = 0.0892$

$P (X = 4) = \frac{e^{- 6} \times 6^{4}}{4!} = 0.1339$ $P(X=4) = (e^-6 xx6^4)/(4!) = 0.1339$

$P (X = 5) = \frac{e^{- 6} \times 6^{5}}{5!} = 0.1606$ $P(X=5) = (e^-6 xx6^5)/(5!) = 0.1606$

$P (X = 6) = \frac{e^{- 6} \times 6^{6}}{6!} = 0.1606$ $P(X=6) = (e^-6 xx6^6)/(6!) = 0.1606$

Then

$P (2 < X < 7)$ $P(2 < X < 7)$

$= P (X =3) + P (X =4) + P (X =5) + P (X =6)$ $= P(X"=3")+P(X"=4")+P(X"=5")+P(X"=6")$

$= 0.0892 + 0.1339 + 0.1606 + 0.1606$ $= 0.0892+0.1339+0.1606+0.1606$

$= 0.5443$ $= 0.5443$
$= 54.43 %$ $=54.43%$

Science

Math

Humanities

... and beyond

How do I Find P ( 2 < X < 7 )? (Stats)

I Got the part A, but don't get how to do part B.

1 Answer

Explanation:

Related questions

Impact of this question