31 is two standard deviations below the mean, since
31 = 35 - 4 = 35 - 2×2
33 is one standard deviation below the mean, since
33 = 35 - 2
So, we have
[tex]P(31 < X < 33) = P(-2 < Z < -1)[/tex]
where [tex]X[/tex] has the given normal distribution (mean 35, s.d. 2) and [tex]Z[/tex] has the standard normal distribution (mean 0, s.d. 1).
Recall the empirical (68-95-99.7) rule, which says
- [tex]P(-1 < Z < 1) \approx 68\%[/tex]
- [tex]P(-2 < Z < 2) \approx 95\%[/tex]
- [tex]P(-3 < Z < 3) \approx 99.7\%[/tex]
The normal distribution is also symmetric about its mean, so
- [tex]P(0 < Z < 1) = P(-1 < Z < 0) \approx 34\%[/tex]
- [tex]P(0 < Z < 2) = P(-2 < Z < 0) \approx 47.5\%[/tex]
- [tex]P(0 < Z < 3) = P(-3 < Z < 0) \approx 49.85\%[/tex]
Now,
[tex]P(-2 < Z < -1) = P(-2 < Z < 0) - P(-1 < Z < 0) \approx 47.5\% - 34\% = 13.5\%[/tex]
so the 351 values between 31 and 33 make up about 13.5% of the total body of data in the distribution. Then there must be
[tex]\dfrac{351}{13.5\%} = \dfrac{351}{0.135} = \boxed{2600}[/tex]
values in the whole data set. (A)
