Question

According to a study conducted by the Toronto-based social media analytics firm Sysomos, of all tweets get no reaction. That is, these are tweets that are not replied to or retweeted (Sysomos website, January ). Suppose we randomly select tweets. a. What is the expected number of these tweets with no reaction (to the nearest whole number)? b. What are the variance (to 2 decimals) and standard deviation (to 4 decimals) for the number of these tweets with no reaction? Variance Standard deviation

Answer 1

Complete Question

According to a study conducted by the Toronto-based social media analytics firm Sysomos, 75% of all tweets get no reaction. That is, these are tweets that are not replied to or retweeted (Sysomos website, January ). Suppose we randomly select tweets.

a. What is the expected number of these tweets with no reaction (to the nearest whole number)?

b. What are the variance (to 2 decimals) and standard deviation (to 4 decimals) for the number of these tweets with no reaction? Variance Standard deviation

Answer:

a

The mean is  $\mu = 150$

The variance $\sigma^2 =37.5$

The standard deviation is $\sigma \approx 6.1237$

Step-by-step explanation:

A binomial distribution is mathematically represented as $B(n,p)$

where n denotes the number of independent trials  =200

    And p denote the probability of success  = 0.75

     We also have the probability of failure which is denoted as q and is equivalent to $1-p$

if p and q are constant then

In binomial distribution the probability of x success and (n-x) is mathematically represented as

           $P[X =x] =[\left n} \atop x}} \right. ] p^x (1-p)^{n-x}$

Where x= 0,1,2,...,n $for 0\le p \le 1$

Now $[\left n} \atop x}} \right. ] = \frac{n!}{x(n-x)!}$

Now the mean(Expected number ) of these probability in binomial distribution is mathematically represented as

                  $\mu = np$

                     $= 200 *0.75 = 150$

And the variance is mathematically represented as

                 $\sigma^2 = np(1-p)$

                    $= 200 * 0.75 (1-0.75) = 37.5$

And the standard deviation is mathematically represented as

                     $\sigma = \sqrt{np(1-p)}$

                       $= \sqrt{200 *0.75(1-0.75)} =\sqrt{37.5} \approx 6.1237$

We can obtain the standard z -score of a d