Answer the following.

Question

Assuming that the heights of female college students are normally distributed with mean 64 inches and standard deviation 2.3 inches, answer the following questions. (Hint: Use the figure below with mean 𝜇 and standard deviation 𝜎.)

Area Under

(a)

What percentage of female college students are taller than 64 inches?

 %

(b)

What percentage of female college students are shorter than 64 inches?

 %

(c)

What percentage of female college students are between 61.7 inches and 66.3 inches?

 %

(d)

What percentage of female college students are between 59.4 and 68.6 inches?

 %

Answer 1

Hi Tamera,

(a) mu is 64 inches, so just look at the percentages in the area to the right for taller females. Add these together: 34 + 13.5 + 2.35 + 0.15 =50%

(b) It is essentially the same procedure on the left side. Values are identical. 34.5 + 13.5 + 2.35 + 0.15 = 50%

(c) Mean is 64 inches, standard deviation is 2.3 inches, so take: 64 +/- 2.3. This gives (61.7, 66.3). So, those marks are exactly one standard deviation from the mean in either direction. The figure above is basically an illustration of the 68-95-99.7 rule, which states that in normal distributions, 68% of the values are within one standard deviation of the mean, 95% are within two, and 99.7% are within three. You were asked for 61.7 to 66.3 inches, so your answer is 68%.

(d) This is basically the same process as (c), but we are interested in two standard deviations from the mean in either direction, not just one. 64 +/- 2(2.3)= (59.4, 68.6). Recall from above that 95% of data fall within two standard deviations of the mean in normal distributions, so your answer is: 95%.

I hope this helps.