$75, $97, $360, $84, $119, $100 Find the median a. 45 b. 98.5 C. 97 d. 100

You are at a stall at a fair where you have to throw a ball at a target. There are two versions of the game. In the first

Tomtit [17]

Answer:

$P(X=0)=(3C0)(0.1)^0 (1-0.1)^{3-0}=0.729$

And the probability of loss with the first wersion is 0.729

$P(Y=0)=(5C0)(0.05)^0 (1-0.05)^{5-0}=0.774$

And the probability of loss with the first wersion is 0.774

As we can see the best alternative is the first version since the probability of loss is lower than the probability of loss on version 2.

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Alternative 1

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=3, p=0.1)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

We can find the probability of loss like this P(X=0) and if we find this probability we got this:

$P(X=0)=(3C0)(0.1)^0 (1-0.1)^{3-0}=0.729$

And the probability of loss with the first wersion is 0.729

Alternative 2

Let Y the random variable of interest, on this case we now that:

$Y \sim Binom(n=5, p=0.05)$

The probability mass function for the Binomial distribution is given as:

$P(Y)=(nCy)(p)^y (1-p)^{n-y}$

Where (nCx) means combinatory and it's given by this formula:

$nCy=\frac{n!}{(n-y)! y!}$

We can find the probability of loss like this P(Y=0) and if we find this probability we got this:

$P(Y=0)=(5C0)(0.05)^0 (1-0.05)^{5-0}=0.774$

And the probability of loss with the first wersion is 0.774

As we can see the best alternative is the first version since the probability of loss is lower than the probability of loss on version 2.

4 0

4 years ago

The diameter of a circle is 10 in. Find its area in terms of pi

elena55 [62]

Answer:

25 pi

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

An instructor who taught two sections of engineering statistics last term, the first with 25 students and the second with 35, de

Amiraneli [1.4K]

Answer:

a) P=0.1721

b) P=0.3528

c) P=0.3981

Step-by-step explanation:

This sampling can be modeled by a binominal distribution where p is the probability of a project to belong to the first section and q the probability of belonging to the second section.

a) In this case we have a sample size of n=15.

The value of p is p=25/(25+35)=0.4167 and q=1-0.4167=0.5833.

The probability of having exactly 10 projects for the second section is equal to having exactly 5 projects of the first section.

This probability can be calculated as:

$P=\frac{n!}{(n-k)!k!}p^kq^{n-k}= \frac{15!}{(10)!5!}\cdot 0.4167^5\cdot0.5833^{10}=0.1721$

b) To have at least 10 projects from the 2nd section, means we have at most 5 projects for the first section. In this case, we have to calculate the probability for k=0 (every project belongs to the 2nd section), k=1, k=2, k=3, k=4 and k=5.

We apply the same formula but as a sum:

$P(k\leq5)=\sum_{k=0}^{5}\frac{n!}{(n-k)!k!}p^kq^{n-k}$

Then we have:

$P(k=0)=0.0003\\P(k=1)=0.0033\\P(k=2)=0.0165\\P(k=3)=0.0511\\P(k=4)=0.1095\\P(k=5)=0.1721\\\\P(k\leq5)=0.0003+0.0033+0.0165+0.0511+0.1095+0.1721=0.3528$

c) In this case, we have the sum of the probability that k is equal or less than 5, and the probability tha k is 10 or more (10 or more projects belonging to the 1st section).

The first (k less or equal to 5) is already calculated.

We have to calculate for k equal to 10 or more.

$P(k\geq10)=\sum_{k=10}^{15}\frac{n!}{(n-k)!k!}p^kq^{n-k}$

Then we have

$P(k=10)=0.0320\\P(k=11)=0.0104\\P(k=12)=0.0025\\P(k=13)=0.0004\\P(k=14)=0.0000\\P(k=15)=0.0000\\\\P(k\geq10)=0.032+0.0104+0.0025+0.0004+0+0=0.0453$

The sum of the probabilities is

$P(k\leq5)+P(k\geq10)=0.3528+0.0453=0.3981$

8 0

3 years ago

Triangle ABC is dilated about the origin with a scale factor of 3.

vitfil [10]

Answer:(0)-12,-15

Step-by-step explanation:

4 0

3 years ago

A triangle has angles measuring 30 degrees and 90 degrees the length of the included side is 6cm. tell whether the conditions fo

Alexxx [7]

Angle 1 = 30°
Angle 2 = 90°
⇒ Angle 3 = 60°
So, it's right triangle. We can set the length of one side and get all other sides.
So, we have 1 triangle. If there are 2 or more triangles with the same data, all the triangles will be congruent because of : Two triangles are congruent if "ASA: Two interior angles and the included side in a triangle have the same measure and length, respectively, as those in the other triangle."

4 0

3 years ago