Brianna has 3 posters she wants to hang on the wall. How many different ways can

At a financial institution, a fraud detection system identifies suspicious transactions and sends them to a specialist for revie

labwork [276]

Answer:

a. E(X) = 54.4

b. E(X) = 2.5

c. P(Y=2) = .0116

Step-by-step explanation:

a.

E(X) = np = .40 probability * 136 trials = 54.4 blocked transmissions

To get the expected value, we simply multiply probability times number of trials. You can look at it in simple terms by thinking if there's a 50% chance of flipping heads and you flip a coin twice, in an ideal world you will have .5*2 = 1 head.

b.

i. Let X represent the number of suspicious transmissions reviewed until finding the first blocked one. We will use a geometric distribution to model the "first" transmission. Whenever we're looking for the "first" time something happens, we use geometric.

ii. E(X) = 1/p , according to the geometric model.

= 1/.4 = 2.5.

We expect that the specialist will review 2.5 suspicious transactions <em>on average </em>before finding the first transmission that will be blocked.

c.

i. Let Y represent the exact number of blocked transmissions out of 10. We will use a binomial distribution to model the "fixed" number of transmissions. Whenever we're looking for a "fixed" number of times something happens, we use binomial.

ii. P(Y=k) = (n choose k)(p^k)(q^n-k)

P(Y=2) = (¹⁰₂)(.4^2)(.6^10-2)

= 45 (.4^2)(.6^10-2) = .0016

As for calculator notation, the n choose k can be accessed on a TI-84 via MATH -> PRB -> nCr. It looks like 10 nCr 2 on the display.

Hence the probability that two transactions out of ten will be blocked is .0016 by the binomial model.

5 0

3 years ago

J is the Midpoint of Segment MN. If MJ=23, Find MN.

Helga [31]

Answer:

46?

Step-by-step explanation:

7 0

3 years ago

HELP <br> Write an equation that represents the line.<br> Use exact numbers.

horrorfan [7]

Y=7/5+6.8

I believe it can a little more than 6.8 but either way this is correct

3 0

2 years ago

The weight of an adult swan is normally distributed with a mean of 26 pounds and a standard deviation of 7.2 pounds. A farmer ra

Snezhnost [94]

Let $X$ denote the random variable for the weight of a swan. Then each swan in the sample of 36 selected by the farmer can be assigned a weight denoted by $X_1,\ldots,X_{36}$ , each independently and identically distributed with distribution $X_i\sim\mathcal N(26,7.2)$ .

You want to find

$\mathbb P(X_1+\cdots+X_{36}>1000)=\mathbb P\left(\displaystyle\sum_{i=1}^{36}X_i>1000\right)$

Note that the left side is 36 times the average of the weights of the swans in the sample, i.e. the probability above is equivalent to

$\mathbb P\left(36\displaystyle\sum_{i=1}^{36}\frac{X_i}{36}>1000\right)=\mathbb P\left(\overline X>\dfrac{1000}{36}\right)$

Recall that if $X\sim\mathcal N(\mu,\sigma)$ , then the sampling distribution $\overline X=\displaystyle\sum_{i=1}^n\frac{X_i}n\sim\mathcal N\left(\mu,\dfrac\sigma{\sqrt n}\right)$ with $n$ being the size of the sample.

Transforming to the standard normal distribution, you have

$Z=\dfrac{\overline X-\mu_{\overline X}}{\sigma_{\overline X}}=\sqrt n\dfrac{\overline X-\mu}{\sigma}$

so that in this case,

$Z=6\dfrac{\overline X-26}{7.2}$

and the probability is equivalent to

$\mathbb P\left(\overline X>\dfrac{1000}{36}\right)=\mathbb P\left(6\dfrac{\overline X-26}{7.2}>6\dfrac{\frac{1000}{36}-26}{7.2}\right)$
$=\mathbb P(Z>1.481)\approx0.0693$

5 0

3 years ago

Please help me with 5

iren2701 [21]

Answer:

Fourth chice, 3/5

Step-by-step explanation:

Add all the numbers together, 12 + 3 + 5 = 20

There's 12 green cubes out of the 20 cubes so it's 12/20 which is 3/5

5 0

3 years ago