In Data Patterns – III we discussed z score and the area under the normal curve table. Let us now look at some practical problems using the table.
Problem 1: 500 people were asked to switch off all electric appliances when they left a room. They were observed for 3 months. It was found that “normally” people did turn off all appliances about 40 days out of the 3 months, give or take 8 days. If this is the case, what percentage of the people would have done it between 36 and 48 days?
Sample size, N = 500
Mean of the sample, μ = 40
Standard deviation, σ = 8
The z score tells us how many standard deviations a data point is away from the mean.
Using the formula (given in data patterns – III), for 36, z score = – 0.5. So, 36 is 0.5 standard deviation away from the 40 days mark, and, because of the “-” sign, we know it is below the 40 days mark. Of course, it’s quite obvious that 36 < 40, but for thoroughness, we explained the “-” sign.
Similarly, the z score for 48 is 1. So, 48 is 1 standard deviation away from the 40 days mark, and because the sign is not “-“, we know it is above the 40 days mark. Once again, it’s obvious that 48 > 40, but for thoroughness, we had to explain the lack of “-” sign.
Now, let’s pull up the table.
Let’s start with 36. The z score for 36 is – 0.5.
0.5 is matched with the value 1915 in the table.
This means that 19.15% of the points lie between 36 and the mean (40).
So, 19.15% of the people did the task between 36-40 days.
Now let’s do 48. The z score for 48 is 1.
1 is matched with the value 3413 in the table.
This means 34.13% of the points lie between 48 and the mean (40).
So, 34.13% of the people did the task between 40-48 days.
Therefore, the percentage of people who did the task between 36 and 48 days is 19.15 + 34.13 = 53.28%.
So, a little over half the people did the task for 36-48 days out of the entire 3 month period. I’d be pretty disappointed by that. It means the sample clearly did not take the task very seriously. But there you are.
Problem 2: 75 cocaine addicts were asked to deny cocaine in exchange for 500 dollars in a 100 minutes window. It was found that most of them managed about 50 minutes before giving in, give or take 10. What was the time range that 60% of the participants managed?
Sample size, N = 75
Mean of the sample, μ = 50
Standard deviation, σ = 10
We know that approximately 68% of the data points lie within 1 standard deviation of the mean. The way we got that is by taking 34% on each side of the mean and then doubling that (if this bit is unclear, please take a minute to revisit the last section of Data patterns – III).
Using similar logic, we can say that we’re looking now for 30% on each side of the mean, which means we’re looking for the value 3000 in the table.
Look back at the table. You’ll see that there is no 3000. But there is a 2995 (29.95%) at 0.84, which is close enough.
So, to account for 30% of the data points, the mark should be 0.84 standard deviations away from the mean (50 minutes).
So we’re talking about z scores of +0.84 (for 30% above the mean) and -0.84 (for 30% below the mean).
Now, in the formula, z = (x – μ)/σ, we know the values of z, μ and σ. We just need to find x for both z = +0.84 and z = -0.84.
Doing so, we get x = 41.6 (for -0.84) and x = 58 (for +0.84).
Therefore, the time range that 60% (30% + 30%) of the participants managed is (41.6 – 58) minutes out of the 100 minute window that was given.
Thank you for reading.