Assume I just completed a study of how many hours per day teenagers between the ages of 14 and 16 spend on their mobile phone per day. I randomly chose 100 subjects. It turns out my data indicated that they spent on average 3.2 hours per day on their phone. The standard deviation was 0.5 hours. If the data is normally distributed:

how many teenagers spent between 3.2 hours and 3.7 hours on their mobile phone?

To calculate the number of teenagers who spent between 3.2 hours and 3.7 hours on their mobile phone, we need to calculate the z-scores for both values.

The formula for calculating the z-score is:

z = (x - μ) / σ

Where:

z is the z-score
x is the value we want to find the probability for (in this case, 3.2 and 3.7 hours)
μ is the mean (3.2 hours)
σ is the standard deviation (0.5 hours)

Let's calculate the z-scores:

For 3.2 hours:

z = (3.2 - 3.2) / 0.5 = 0

For 3.7 hours:

z = (3.7 - 3.2) / 0.5 = 1

Now, we need to find the probabilities associated with these z-scores using a z-table.

Looking up the z-score of 0 in the z-table, we find that the corresponding probability is 0.5000.

Looking up the z-score of 1 in the z-table, we find that the corresponding probability is 0.8413.

To find the number of teenagers who spent between 3.2 hours and 3.7 hours on their mobile phone, we subtract the probability associated with the lower z-score from the probability associated with the higher z-score:

Number of teenagers = 0.8413 - 0.5000 = 0.3413

Therefore, approximately 34.13% of the teenagers spent between 3.2 hours and 3.7 hours on their mobile phone.

Learn more about calculating the number of teenagers who spent a specific amount of time on their mobile phone here:

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