Section 3 answer key:
1) The probability density curve is a rectangle from 5 to 21, which is 21-5=16 long. Since the area must be 1, (16)h=1,
and the height is 1/16=0.0625. If X is the number we get, the probabilities are represented by areas above the intervals.
(a) X>18: The rectangle above (18,21) is 3 long and 1/16 high, so 3/16.
(b)
: The rectangle above [15,20] is 5 long and 1/16 high, so 5/16.
(c)
: The rectangle above (15,20) is 5 long and 1/16 high, so 5/16. The presence or absence of exactly 15
and/or exactly 20 contributes no area.
(d) Exactly 13. There is no area above this point; 0.
(e)
: A point makes this true if it’s in either of the intervals [5,10) or (17,21]. The total area above
these two intervals is the two rectangles which are 10-5=5 and 21-17=4 long respectively. The two areas are 5/16 and
4/16, so the total area is 9/16.
(f)
: A point makes this true if it’s in either of the intervals [5,17) or (10,21]; that is, every point
makes this true. The total area above these two intervals is the entire rectangle, which has area 1. Making both conditions
true (e.g., X is 14) doesn’t make it any more true; no area is double-counted.
(g) The mean is the “balance line” of the rectangle, which is obviously its midpoint: (5+21)/2=13. The median is the
line where half the area is on either side, which is (for a rectangle) obviously the same place. The quartiles are the lines
where a quarter and three-quarters of the area is on either side. For a rectangle, this is obviously one-quarter and threequarters of the way across, so two more midpoints (5+13)/2=9 and (13+21)=17. Alternately, we’re dividing the original
length 16 into four equal intervals of length 16/4=4, so add it to 5, add again, add again.
2) The isosceles triangle above [0,1] has length 2 and area 1. Its height h can then be found: ½*b*h= ½ * (1)*h=1, so
h=2. The probability that Y is less than ¼ is the area above the interval [0,1/4]. This is a smaller triangle which is similar
to the larger triangle above [0,1/2], a right triangle with length ½ and height 2. Setting up a proportion for similar
triangles, the height h of the small triangle satisfies:
Alternately, if we halve the base, we are going to halve the height.
Since this triangle has length 0.25 and height 1, the area is ½*0.25*1=0.125 or 1/8.
Note that for a rectangle (1), a quarter of the horizontal distance gives a quarter of the area above. For a triangle (or
anything that peaks in the middle), this won’t be the case. The smallest quarter [0,1/4] occurs 1/8 of the time. Likewise
for the biggest quarter [3/4,1], by symmetry. The horizontal middle [1/2, ¾] thus occurs the remaining 6/8 of the time.
That is, the number is three times more likely to be in the middle half (6/8) than it is to be in the outlying half (2/8).
Numerically, we can see that a small “average of two spins” is possible, but unlikely; for Y to be less than ¼, we must get
two small spins in a row. This is possible, but not just as likely as everything else. It’s more likely that one small spin
and one large spin will average up to a “central” Y.
3) X=N(266,16)
(a) The middle 68% of all pregnancies last between 266-16 and 266+16 days, 250 to 282. The middle 95% of all
pregnancies last between 266-2*16 and 266+2*16 days, 234 to 298 (for future reference, note that this “rule” is rounded
somewhat compared to the charts). The middle 99.7% of all pregnancies last between 266-3*16 and 266+3*16 days, 218
to 314.
c) We wish to find the tick-mark which separates 2.5% of the area (0.0250) to the left. The closest entry (exact, in fact) is
z= 1.96. Multiply by 16 and add 266 to get 234.54. We wish to find the tick-mark which separates 2.5% of the area
(0.0250) to the right, or 0.9750 to the left. The closest entry (exact, in fact) is z=+1.96. Multiply by 16 and add 266 to get
297.36. In passing, notice that this is another way of asking for the tick-marks that separate out the middle 95%.
Compare with (a); these answers are slightly more accurate than the “rule”.
d) We wish to find the tick-mark which separates 25% of the area (0.2500) to the left. The closest entry is z= 0.67.
Multiply by 16 and add 266 to get 255.28. We wish to find the tick-mark which separates 25% of the area (0.2500) to the
right, or 0.7500 to the left. The closest entry is z=0.67. Multiply by 16 and add 266 to get 276.72.
e) We wish to find the tick-mark which separates 14% of the area (0.1400) to the left. The closest entry is z=
.
Multiply by 16 and add 266 to get 248.72. We wish to find the tick-mark which separates 14% of the area (0.1400) to the
right, or 0.8600 to the left. The closest entry is z=1.08. Multiply by 16 and add 266 to get 283.28.
4) S=N(500,100) and A=N(18,6). Standardizing their scores:
5) S=N(500,100).
6) X=N(544,103)
We wish to find the tick-mark which separates 10% of the area (0.1000) to the left. The closest entry is z= 1.28.
Multiply by 103 and add 544 to get 412.16. We wish to find the tick-mark which separates 10% of the area (0.1000) to
the right, or 0.9000 to the left. The closest entry is z=1.28. Multiply by 103 and add 544 to get 675.84.
7.) X=N(3.8, 0.2)
8)
While all are exceptionally good (off the chart), 4.26 sweeps out the most area to the left.
c) We wish to find the tick-mark which separates 95% of the area (0.9500) to the left. The closest entries are z=1.64 or
1.65. Multiply by 0.0317 and add 0.261 to get about 0.313 in either case.
d) We wish to find the tick-marks which separate 30% and 70% of the area (0.3000 and 0.7000) to the left. The closest
entries are z= 0.52 and +0.52. Multiply by 0.0371 and add 0.266 to get 0.247 and 0.285
9) X=N(224, 0.03)
10) X=N(188,24)
c) We wish to find the tick-mark which separates 85% of the area (0.8500) to the left. The closest entry is z=1.04.
Multiply by 24 and add 188 to get 212.96
11) X=N(32000,2500). What is the probability of a single tire lasting at least 40,000 miles?
The probability of a particular tire lasting this long is very small; the probability of getting four such exceptional tires in a
row must be very, very small (see section 4 for details).
d) We wish to separate out the worst 1/25 (shortest treadlives), that is, to find the tick-mark which separates 4% of the
area (0.0400) to the left. The closest entry is z= 1.75. Multiply by 2500 and add 32000 to get 27,625 miles.
12) X=N(38.2,1.8)
b) We wish to find the tick-marks which separate 10% and 90% of the area (0.1000 and 0.9000) to the left. The closest
entries are z= 1.28 and +1.28. Multiply by 1.8 and add 38 to get 35.696 and 40.304 inches.
c) This is the same question as (b), so 40.304 inches.
13) X=N(98.2, 0.7)
b) We wish to find the tick-mark which separates 20% (0.2000) to the left. The closest entry is z=
and add 98.2 to get 97.612 degrees.
. Multiply by 0.7
d) We wish to find the tick-marks which separate 35% and 65% of the area (0.3500 and 0.6500) to the left. The closest
entries are z=
and +0.39. Multiply by 0.7 and add 98.2 to get 97.927 and 98.473 degrees.