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Vsevolod [243]
2 years ago
10

According to a study conducted by the Toronto-based social media analytics firm Sysomos, 71% of all tweets get no reaction. That

is, these are tweets that are not replied to or retweeted (Sysomos website, January 5, 2015). Suppose we randomly select 100 tweets. What is the expected number of these tweets with no reaction (to the nearest whole number)
Mathematics
1 answer:
timama [110]2 years ago
3 0

Answer:

Let X the random variable of interest "number of tweets with no reaction", on this case we now that:

X \sim Binom(n=100, p=0.71)

And the expected value is given by:

E(X) = np =100*0.71 = 71

So we expect about 71 tweets with no reaction for this case.

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=\frac{n!}{(n-x)! x!}  

Solution to the problem

Let X the random variable of interest "number of tweets with no reaction", on this case we now that:

X \sim Binom(n=100, p=0.71)

And the expected value is given by:

E(X) = np =100*0.71 = 71

So we expect about 71 tweets with no reaction for this case.

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The length of time it takes to find a parking space at 9 A.M. follows a normal distribution with a mean of 7 minutes and a stand
julsineya [31]

Answer:

There is a 2.28% probability that it takes less than one minute to find a parking space. Since this probability is smaller than 5%, you would be surprised to find a parking space so fast.

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 = \frac{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.

Also, a probability is unusual if it is lesser than 5%. If it is unusual, it is surprising.

In this problem:

The length of time it takes to find a parking space at 9 A.M. follows a normal distribution with a mean of 7 minutes and a standard deviation of 3 minutes, so \mu = 7, \sigma = 3.

We need to find the probability that it takes less than one minute to find a parking space.

So we need to find the pvalue of Z when X = 1

Z = \frac{X - \mu}{\sigma}

Z = \frac{1 - 7}{3}

Z = -2

Z = -2 has a pvalue of 0.0228.

There is a 2.28% probability that it takes less than one minute to find a parking space. Since this probability is smaller than 5%, you would be surprised to find a parking space so fast.

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2 years ago
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