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According to a recent research, internet users spend an average of 2.3 hours per day on social media, with a standard deviation of .3. It is thought that internet usage follows a normal distribution. A random sample of 6 internet users is selected.

According to a recent research, internet users spend an average of 2.3 hours per day-example-1
User JSArrakis
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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
User Aanrv
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