An online game site surveys 50 random players to find out how many times per week the user plays any video game online. the mean for the site's data is 154. thomas decides to do his own study. he asks the first 50 people who enter the lunchroom how many video games they play each week. he determines that the mean for his data is 20. what can thomas do so that his statistical results are more in line with the statistical results of the online site?

When the online site surveyed about how many times per week the user plays any video game online, the mean was 154. When Thomas surveyed individually to first 50 persons, he got the mean 20.

Now, if Thomas wants that his statistical results should be more accurate and inline than the statistical results, he need to increase the samples Like 100-200 people. As the sample increases, the statistical results are more accurate.

a. 12.5 kg

step-by-step explanation:

f = ma

so

m = f/a = (2500 n)/(200 m/s²) = 12.5 kg

1 n = 1 kg·m/s²

answer:   35 % is the probability of a late departure or an early arrival.

step-by-step explanation:

let l represents the late departure and e represents the early departure.

then, according to the question,

p(l) = 12 %,   p(e) = 27 % and = 4 %

and, we have to find out,

but, we know that,

which is the required equation that will use to find the probability of a late departure or an early departure.

therefore,   the probability of a late departure or an early arrival =   35 %