Researchers doing a study comparing time spent on social media and time spent on studying randomly sampled 200 students at a major university. They found that students in the sample spent an average of 2.3 hours per day on social media and an average of 1.8 hours per day on studying. If all the students at the university in fact spent 2.2 hours per day on studying, with a standard deviation of 2 hours, the probability of getting a sample average of 1.8 or less is:.a. 0.5893 b. 0.9977 c. 0.4207 d. 0.0023

d. 0.0023

Step-by-step explanation:

Calculation for the probability of getting a sample average of 1.8 or less

Using this formula

z score =(X-mean)/Standard errror

Where,

Standard error =Standard deviation/√(n)

Let plug in the formula

P(Xbar<1.8)=P(Z<(1.8 hours per day -2.2 hours per day)/(2 hours /√(200))

P(Xbar<1.8)=P(Z<-0.4/(2/14.14))

=P(Z<-0.4/0.1414)

=P(Z<-2.83)

Now let find the Z score of P(Z<-2.83) using the z table

P=0.0023

Therefore the probability of getting a sample average of 1.8 or less is:0.0023

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step-by-step explanation:

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