Michael M. answered • 11/23/22
Data Scientist with over 12 Years Instruction Experience
a) When determining something like this, it's helpful to remember the definitions for standard deviation and z-scores. Remember, the z-score is, essentially, how many standard deviations away from the mean a target number is. Then, the z-score lookup tables allow us to "look up" the percentage of a population that falls within our desired range (in this case, all people above 95 IQ).
First, we calculate the z-score by figuring out "how many standard deviations away from the population mean (100) is 95?" To do this, we figure out the different between the two numbers before dividing that difference by the standard deviation. (So, with a standard deviation of 15, if we were trying to figure out the z-score of something like 115, it would have a z-score of 1 as it's exactly 1 standard deviation away from the mean). To calculate this, use the basic formula for z-score calculations: z=(x-µ)/σ where x is the number we're figuring out the z-score for (95), µ is the population mean (100), and σ is the population standard deviation (15). So, for this calculation, we get a z-score of -0.3333.
Then, you look up this value in a z-score table. When you find the value, remember that most z-score tables show you the left-tailed test probability (meaning the value probability to the left of the test value, 95 in our case). However, we want to determine the probability that an IQ is greater than 95, so it's a right-tailed test. To convert the left-tailed value to a right-tailed value, subtract the left-tailed value from 1. This should get you to a value just above 0.6 (I'll let you find it and convert to %).
b) Same thing as before, but now our value is 125 and we're doing a left-tailed test. Same concept, we first calculate the z-score then look that value up in a z-score table. This time, though, we won't have to do any conversions as the z-score table values show the left-tailed test probability.
125-100 = 25
25/15 = 1.6667 (this is your z-score. 125 is one and two-thirds standard deviations away from the mean)
Look this value up in the table and convert to a percentage.
c) Same general problem as (b), as you're doing a left tailed test (with 110 instead of 125). The only difference here is that the problem is asking you to use the percentage you find to calculate the number of people in 600 that meets that criterion. (Follow same steps to calculate z-score, and remember, it's a left-tailed test so now conversion is required after you look up the value. Just use that value to calculate how many people in 600 would have IQs less than 600).
d) Same general problem as (a), as it's a right-tailed test (with 140 instead of 95). Follow the same steps as (a) until you get the percentage, then do the same thing you did in (c) to get the number of people.
Good luck!
