Dan K. answered • 10/07/23
Northwestern University PhD specializing in psych, stats, and R
Hi Bailey,
The middle X% of a normal distribution always corresponds to how many standard deviations (SDs) above and below the mean you’re dealing with. For example, we know from the empirical rule that the middle 68% of the normal distribution falls between 1 SD below and 1 SD above the mean. In this problem, since 1 SD = 14 days, that means that 68% of women have pregnancies that are between 14 days shorter than the mean (263) and 14 days longer--i.e., 68% of women have pregnancies that are between 249 and 277 days long.
But this question isn’t asking about the middle 68%, it’s asking about the middle 98%. So if we know that 68% of the normal distribution corresponds to ± 1 SD from the mean (or, equivalently, ± 14 days from the mean), how many SDs above and below the mean correspond to the middle 98% of the distribution?
Well, another way to talk about SDs when it comes to the normal distribution is to use the term Z-score. A Z-score of +1 corresponds to 1 SD above the mean, while a Z-score of -1 corresponds to 1 SD below the mean. So 68% of the normal distribution falls between Z = -1 and Z = 1.
To figure out the Z-scores that define the middle 98% of the distribution, we need to use a pre-calculated Z-table or an online calculator. This calculator (Z-score Calculator) includes handy graphs and shows us that the middle 98% of the distribution falls between Z = -2.326 and Z = 2.326.
If a Z-score of 1 corresponds to 1 SD and therefore, in this problem, 14 days, then a Z-Score of 2.326 corresponds to 2.326 SDs, and therefore 2.326 * 14 = 32.564 days.
Now we have all the pieces we need to find our range! 98% of pregnancy lengths will fall between the mean and ± 2.326 SDs.
That is, 263 ± 32.564 days, or between 230 and 296 days, rounded to the nearest one’s place.
