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inysia [295]
3 years ago
6

According to a 2010 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. a). What is the expected number of these tweets with no reaction? b). What are the variance and standard deviation for the number of these tweets with no reaction?
Mathematics
1 answer:
klemol [59]3 years ago
6 0

Answer:

a) the expected number of tweets with no reaction is 71 tweets

b) the variance is 20.59 tweets² ( ≈ 21 tweets )and the standard deviation is

4.537 tweets (≈ 5 tweets)

Step-by-step explanation:

we can use the binomial probability distribution

P(x, N, p ) = N!/[(N-x)!x!] * p^x * (1-p)^(N-x)

where

x = number of successful events

N = population total

p = probability for success for every individual and independent event

P = probability for x successful events

in our case p = 71% = 71/100 (probability of a tweet without reaction) , N = 100 tweets

a) the expected number of tweets

E(x) = N* p = 100 tweets * 71/100 = 71 tweets

b) the variance is

V(x) = N * p * (1-p)

V = 100 * 71/100 * (1-71/100) = 71* 29/100 = 20.59 tweets² ( units of variance are [N²] )

the standard deviation is

s = √V = √20.59 tweets² = 4.537 tweets

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valentinak56 [21]

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Give a 98% confidence interval for one population mean, given asample of 28 data points with sample mean 30.0 and sample standar
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The confidence interval for the population mean, given the sample mean μ and the sample standard deviation σ, can be calculated as:

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Where n is the sample size, and Z is the z-score for 1 - α/2. Using the known values:

CI(\mu)=\lbrack30.0-Z_{0.99}\cdot\frac{2.40}{\sqrt[]{28}},30.0+Z_{0.99}\cdot\frac{2.40}{\sqrt[]{28}}\rbrack

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