Select a discrete probability distribution and present a real-life application of that distribution. Interpret the expected value and the standard deviation of your selected distribution within the context of the real-life example that you have selected, and describe how these values can be used by enterprise decision-makers.

Answers

Answer 1
Answer:

Answer:

If a new product wants to be tested by a company and decides to show 50 samples of this product to 50 selected customers. The company estimates that the probability that the customer buys the product is 0.67, the objective is to determine approximately how many people expect to buy the product.

Let X the random variable of interest "Number of people that will buy a selected product", on this case we now that:  

X \sim Binom(n=50, p=0.67)  

The expected value is given by this formula:

E(X) = np=50*0.67=33.50

And the standard deviation for the random variable is given by:

sd(X)=√(np(1-p))=√(50*0.67*(1-0.67))=3.32

So then they can conclude that for each group of 50 people they expect that about 33-34 peoploe will buy the product with a standard deviation of 3.32.

Step-by-step explanation:

Previous concepts

A Bernoulli trial is "a random experiment with exactly two possible outcomes, "success" and "failure", in which the probability of success is the same every time the experiment is conducted". And this experiment is a particular case of the binomial experiment.

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".

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=(n!)/((n-x)! x!)  

Solution to the problem

If a new product wants to be tested by a company and decides to show 50 samples of this product to 50 selected customers. The company estimates that the probability that the customer buys the product is 0.67, the objective is to determine approximately how many people expect to buy the product.

Let X the random variable of interest "Number of people that will buy a selected product", on this case we now that:  

X \sim Binom(n=50, p=0.67)  

The expected value is given by this formula:

E(X) = np=50*0.67=33.50

And the standard deviation for the random variable is given by:

sd(X)=√(np(1-p))=√(50*0.67*(1-0.67))=3.32

So then they can conclude that for each group of 50 people they expect that about 33-34 peoploe will buy the product with a standard deviation of 3.32.


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The time that it takes a randomly selected job applicant to perform a certain task has a distribution that can be approximated by a normal distribution with a mean value of 145 sec and a standard deviation of 25 sec. The fastest 10% are to be given advanced training. What task times qualify individuals for such training? (Round the answer to one decimal place.)

Answers

Answer:

A task time of 177.125s qualify individuals for such training.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by

Z = (X - \mu)/(\sigma)

After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.

In this problem, we have that:

A distribution that can be approximated by a normal distribution with a mean value of 145 sec and a standard deviation of 25 sec, so \mu = 145, \sigma = 25.

The fastest 10% are to be given advanced training. What task times qualify individuals for such training?

This is the value of X when Z has a pvalue of 0.90.

Z has a pvalue of 0.90 when it is between 1.28 and 1.29. So we want to find X when Z = 1.285.

So

Z = (X - \mu)/(\sigma)

1.285 = (X - 145)/(25)

X - 145 = 32.125

X = 177.125

A task time of 177.125s qualify individuals for such training.

1. Which fraction has a value that's equal to 7/8​

Answers

Answer:

to find the right answer multiple by 2

Step-by-step explanation:

and you get 14/16

What is (1/3) x (-2/3 )?

Answers

Answer:

-0.2222

Step-by-step explanation:

If its multiple choice choose an answer close to this. Or don't I can't tell what to do.

Answer: -2/9

Decimal form: -0.2

Step-by-step explanation:

Round 34.037 to 3 significant figures.
a. 34.04
b. 34
c. 34.037
d. 34.0​

Answers

Answer:

My answer to the question is 34.0

Round all answers to 4 decimal places.a. A bag contains 4 black marbles, 10 white marbles, and 9 red marbles. If a marble is drawn from the
bag, replaced, and another marble is drawn, what is the probability of drawing first a black marble and
then a red marble?
b. A bag contains 10 blue marbles, 9 red marbles, and 4 white marbles. If two different marbles
are drawn from the bag , what is the probability of drawing first a blue marble and then a white marble?

Answers

a. (4/23)(9/23) = .0681

b. (10/23)(4/23) = .0756

A manufacturer knows that their items have a normally distributed lifespan, with a mean of 8.8 years, and standard deviation of 2.2 years.If 5 items are picked at random, 5% of the time their mean life will be less than how many years?

Answers

First of all we need o use the standard normal table to see what value if z will have 10% of the area (probability) to the left of it. We see that z value is approximately z = -1.282

Now all we have to do is calculate the value of x that corresponds to this value of z. We have

( x - µ ) / σ = -1.282 Substituting in the known values, we have

( x - 4.4 ) / 1.3 = -1.282 and x = 2.73 years