SOLUTION: U.S. internet users spend an average of
18.3 hours a week online. If 95% of users spend between 13.1 and 23.5 hours a week, what is the probability that a randomly selected user i
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Question 1204642: U.S. internet users spend an average of
18.3 hours a week online. If 95% of users spend between 13.1 and 23.5 hours a week, what is the probability that a randomly selected user is online less than 15 hours a week?
Answer by MathLover1(20850) (Show Source): You can put this solution on YOUR website!
To solve this problem, we need to use the properties of the normal distribution.
First, we need to calculate the standard deviation () of the distribution. We know that % of the data lies between and hours.
This range corresponds to the interval from to standard deviations from the mean in a normal distribution (since % of the data in a normal distribution is within standard deviations of the mean).
We can set up the following equations to solve for :
------------------------------
Next, we need to calculate the z-score for hours:
The z-score tells us how many standard deviations an element is from the mean. A z-score of
means that 15 hours is standard deviations below the mean.
Finally, we need to find the probability that a randomly selected user is online less than hours a week. This is the same as finding the area to the left of in the standard normal distribution.
You can find this value in a standard normal distribution table or use a calculator with a normal distribution function.
probabilities for the normal distribution:
endpoint:
mean:
standard deviation:
So, the probability that a randomly selected user is online less than hours a week is approximately , or %.
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