Assignment 14 Query

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course MTH 152

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014. ``q Query 14*********************************************

Question: `q Query problem 13.1.6 freq dist 35 IQ scores class width 5

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Your solution:

There are 50 IQ scores recorded from the survey of tenth graders. The low level we are required to begin with is 91 with nine classes of 5 units each. This gives us categories of 91-95, 96-100, 101-105, 106-110, 111-115, 116-120, 121-125, 126-130, and 131-135. There is 1 student with an IQ in the 91-95 range; 1 of 50 is 2%. There are 3 students with an IQ in the 96-100 range; 3 of 50 is 6%. There are 5 students with an IQ in the 101-105 range; 5 of 50 is 10%. There are 7 students with an IQ in the 106-110 range; 7 of 50 is 14%. There are 12 students in the 111-115 range; 12 of 50 is 24%. There are 9 students in the 116-120 range; 9 of 50 is 18%. There are 8 students in the 121-125 range; 8 of 50 is 16%. There are 3 students in the 126-130 range; 3 of 50, as we have already calculated, is 6%. Finally, there are 2 students in the 131-135 range; 2 of 50 is 4%. This gives us a table of:

91-95 1 1/50 = 2%

96-100 3 3/50 = 6%

101-105 5 5/50 = 10%

106-110 7 7/50 = 14%

111-115 12 12/50 = 24%

116-120 9 9/50 = 18%

121-125 8 8/50 = 16%

126-130 3 3/50 = 6%

131-135 2 2/50 = 4%

confidence rating #$&*: 3

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Given Solution:

`a These numbers might or might not be completely accurate, but the calculation of the relative frequencies as percents follows from the tallies:

91-95 1 1/50 = 2%

96-100 3 3/50 = 6%

101-105 5 5/50 = 10%

106-110 7 7/50 = 14%

111-115 12 12/50 = 24%

116-120 9 9/50 = 18%

121-125 8 8/50 = 16%

126-130 3 3/50 = 6%

131-135 2 2/50 = 4% **

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Self-critique (if necessary): OK

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Self-critique Rating: OK

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Question: `q Query problem 13.1.10 stem and leaf for yards gained by 44 rushers

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Your solution:

In order to create a stem and leaf diagram, we must create a column of stems which include each “prefix” of the numbers included in the chart of net yards gained per game. The leaves of the chart are the “suffixes” of these numbers. For example, if one player gained 36 net yards, the stem would be 3 and the leaf would be 6. When we account for all the yards included in the chart we are given, we create a stem and leaf diagram that looks like this:

0 3 7

1 2 6 8 9

2 2 4 5 8 9 9

3 0 2 3 3 6 6 7 9

4 0 1 2 3 3 5 6 9

5 1 4 5 8 6 0 2 7

7 3 3 9

8 6 8

9 4

10 2

11 2

12 3

confidence rating #$&*: 3

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Given Solution:

`aIn order we get the following:

0 3 7

1 2 6 8 9

2 2 4 5 8 9 9

3 0 2 3 3 6 6 7 9

4 0 1 2 3 3 5 6 9

5 1 4 5 8 6 0 2 7

7 3 3 9

8 6 8

9 4

10 2

11 2

12 3 **

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Self-critique (if necessary): OK

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Self-critique Rating: OK

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Question: `q Query problem 13.1.35 empirical probability distribution for letters of alphabet

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Your solution:

The probability of choosing an A is .08. When we divide this by the total probability of choosing A, E, I, O, or U, (which the book tells us is .385), then we get a probability of .208. The probability of choosing an E is .13. When we divide this by the total probability of choosing A, E, I, O, or U, we get a probability of .338. The probability of choosing an I is .065. Divided by .385, the probability of choosing an I is then .169. The probability of choosing an O is .08. Divided by .385, the probability of choosing an O is, like the final probability of choosing an A, .208. The probability of choosing a U is .03. Divided by .385, the probability of choosing a U is then .078.

confidence rating #$&*: 3

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Given Solution:

`aThe probabilities for A, E, I, O, U are:

.08. .13, .065, .08, .03.

The sum of these probabilities is .385

The probabilities of the letters given a vowel:

.08 / .385 = .208

.13 / .385 = .338

.065 / .385 = .169

.08 / .385 = .208

.03 / .385 = .078

These probabilities total 1. **

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Self-critique (if necessary): OK

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Self-critique Rating: OK

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