Investigation 12

a) We want P(X<70), where X has a normal distribution with mu=75 and sigma=5.  The z-score is -1, and the probability is .1587.

b) The sketches should have the same height/width, but one should be centered at 65 and the other at 75.

c) We want (Y>70), where Y has a normal distribution with mu=65 and sigma=5. The z-score is 1, and the probability is .1587.

d) We want to find k such that P(X<k)=.05, where X has a normal distribution with mu=75 and sigma=5.  The z-score has to be -1.645, so we have (k-75)/5 = -1.645, so k=75-1.645(5) = 66.775.

e) We want P(Y>66.775), where Y has a normal distribution with mu=65 and sigma=5.  The z-score is 0.36, and the probability is .3613.

f) The probability has increased.  This makes sense because by reducing the probability of one type of error (concluding that the additive works when it doesn't), the trade-off is that the probability of the other kind of error (concluding that the additive does not work when it really does) has increased.

g) We want P(W>66.775), where W has a normal distribution with mu=65 and sigma=2.  The z-score is 0.89, and the probability is .1874.  This probability has decreased because with a tighter (narrower) distribution, there's less chance of making an error.

Investigation 13

b) Should be in ballpark of .4375, but answers vary.

c) The answer is 7/16, which is .4375.  To see this, the area of the region where they meet is the area of the square minus the areas of the two
triangles where they don't meet.  This is 60^2 - 1/2 * 45 * 45 - 1/2 * 45 * 45 = 60^2 - 45^2.  The probability is this area divided by the area of the
square, giving 1575/3600, which reduces to 7/16. 

d) The region where they meet is no longer symmetric about the 45degree line.  Because, for example, if Tom arrives at 15 past noon and Mary at 30 past noon, then they meet.  But if Mary arrives at 15 past noon and Tom at 30 past noon, then they do not meet.  The approx probability should be somewhere in the ballpark of .431.

e) The key here is that they meet if T-M is greater than .10 and less than 20.  Now the area of this region is 3600 - 1/2 * 40 * 40 - 1/2 * 50 * 50 = 1550, so the probability is 1550/3600, which reduces to 31/72, which is about .431.

f) The scatterplot should look like random scatter, but this time Tom can not arrive between 45 and 60 minutes after noon, and Mary can not arrive between 0 and 15 minutes after noon.  The region where they meet should be like in a). 

g) The approximate probability should be in the ballpark of .444. 

h) The area where they meet is now 45^2 - 1/2 * 15*15 - 1/2 * 45*45, which is 900, so the probability is 900/45^2 = 900/2025, which is 4/9, or about .4444. 

Investigation 14

a) The scatterplot should look like a circle or ellipse with most points close to the center of (30,30). 

b) Answers will vary but should be in the rough ballpark of .711.

c) The new scatterplot should show points even more tightly concentrated around the center, .and the new approx probability should be in the rough ballpark of .966. This makes sense because the smaller std devs mean more points close to the center, where the two people are more likely to meet.

d) The new scatterplot should show points less tightly concentrated around the center, and the new approx probability should be in the rough ballpark of .355. This makes sense because the different means produce fewer points close to the center, and so the two people are less likely to meet.

e) D-L has a normal distribution with mean is 30-30 = 0 and std dev sqrt(10^2+10^2) = sqrt(200) = 14.14.

f) P(-15 < D-L < 15) turns out to be about .711.

g) The std dev of D-L is now sqrt(5^2+5^2) = sqrt(50) = 7.07, and P(-15 < D-L < 15) is now about .966.

h) The mean of D-L is now 40-20 = 20 and the std dev is back to sqrt(200) = 14.14. P(-15 < D-L < 15) is now about .355.

Investigation 15

b) The distribution of population growths in both East and West are skewed to the right.

c) The five-number summaries are:
                min    Q1    median    Q3    max
East        0.8    5.3    9.65 14.58 26.4
West        0.5 8.6 13.3 22.38 66.3

e) To check for outliers in the East: IQR = 14.58 – 5.3 = 9.28, so 1.5IQR = 13.92. Subtracting this from Q1 gives 5.3 – 13.92 = -8.62. Adding this to Q3 gives 14.58 + 13.92 = 28.5. No values in the East are less than -8,62 or greater than 28.5, so there are no outliers in the East.

To check for outliers in the West: IQR = 22.38 – 8.6 = 13.78, so 1.5IQR = 20.67. Subtracting this from Q1 gives 8.6 – 20.67 = -12.07. Adding this to Q3 gives 22.38 + 20.67 = 43.05. The only outlier in the West is Nevada, with 66.3 percent growth.

f) The center of the distribution is higher in the West, revealing that Western states tended to have higher population growth in the 1990’s than Eastern states did. There is also more spread/variability in the population growths among Western states. Both distributions are skewed to the right, and the only outlier is in the West: Nevada with 66.3% growth in the 1990’s.

Investigation 16

a) The five-number summaries are:
                min    Q1    median    Q3    max
1978        42    58    75            81    95
2003        56    87    91            98    110

b) The 2003 distribution has a larger center than the 1978 distribution, which means that tourists generally have to wait longer for an eruption now than they did 25 years ago.

c) The spread was larger in 1978 than in 2003, so tourists can be more sure about when the next eruption will be now than they were 25 years ago.  In other words, the next eruption time is more predictable now than 25 years ago.

d) The histogram of 1978 inter-eruption times reveals a bi-modality: there is a cluster around 50-60 minutes as well as a larger peak around 75-85 minutes.  This bi-modality does not show up in the 2003 distribution.