A) What is the probability that the average time spent on social media by these 6 users is less than 2 hours?
To answer this question, we first need to calculate the standard error. The formula for calculating it from the standard deviation and sample size is:
SE = SD / sqrt(n)
where
SD = standard deviation = .3
n = sample size = 6
Substituting these values in:
SE = .3 / sqrt(6) ≈ .12247
Next, we calculate the Z-score for 2 hours. The formula for this is:
Z = (x - μ) / SE
where
x = value of interest (2 hours)
μ = population mean (2.3 hours)
Substituting these values in:
Z = (2 - 2.3) / .12247 ≈ -2.45
Finally, we look up a Z-score of -2.45 in a standard normal distribution table or use a calculator with a cumulative normal distribution function to find the corresponding probability.
The probability that a randomly selected score falls below Z=-2.45 is approximately .0071 or 0.71%.
B) What is the probability that an individual user spends more than 3 hours on social media?
In this case, we are looking at an individual user rather than a sample mean so we don't need to calculate standard error.
We can again calculate the z-score using:
Z = (x - μ) / SD
Substituting in:
Z= (3-2.3)/0.3≈ 2.33
Then, once again we look up this Z-score in a standard normal distribution table or use our calculator's cumulative normal distribution function.
The area under curve to the right of Z=2.33 represents the proportion of individuals who spend more than 3 hours on social media.
The total area under curve being 1, subtracting cumulative area to left of positive z from 1 gives us desired probability.
So P(Z>2.33)=1-P(Z