Final grades for the course have been posted in the “Grades” area. These grades were submitted to the CUNYFirst system on Saturday, May 26th. I am not sure when they will be available to you on CUNYFirst — if not immediately, hopefully within a day or two.

I wish you all the very best in your future academic pursuits – it was a pleasure working with you this semester.

Best regards,

Mr. Reitz

- A.) Alice’s score was 94.

In order to find the Z score, we use the Z formula. This formula is:

Z equals X minus MU (mu is the same as the mean) divided by standard deviation.

We are given a mean score of 82 and a standard deviation of 6.

Our X is also given as Alice’s score of 94.

We now plug the numbers in and solve:

94-82/6

*Remember always follow PEMDAS (parenthesis, exponent, multiplication, division, addition and subtraction.)

94-82=12

12/6=2

Alice’s Z score is: **2**

- B.) Bob’s score was 76.

To find Bob’s Z score we use the Z formula as well.

Z equals X minus mu (mu is the same as the mean) divided by standard deviation.

Here our mean and standard deviation are the same but our X is different.

Our mean (which is given) is 82 and our standard deviation (also given) is 6.

Our X is Bob’s score which is 76.

Now we plug our numbers into the formula:

76-82/6

Here PEMDAS should also be used which means we subtract before we divide.

76-82=-6

-6/6=-**1**

- C.) Charlie’s score was 98.

To find Charlie’s Z score we use the Z formula as well.

Z equals X minus mu (mu is the same as the mean) divided by standard deviation.

Here our mean and standard deviation are the same but our X is different.

Our mean (which is given) is 82 and our standard deviation (also given) is 6.

Our X is Charlie’s score which is 98.

Now we plug our numbers into the formula:

98-82/6

Here PEMDAS should also be used which means we subtract before we divide.

98-82=16

16/6=**2.67**

Hope this helps!

]]>Step 1: The first step in solving this problem is using the right formula, the correct formula for this problem would be x-u divided by o. X being the given , U being the average and O being the standard deviation. My iPad does not have the actual symbols but the correct mathematical symbol for U and O has a line at he corner.

Step 2: We take our given numbers from our problem and simple just plug them in. 100-134/20= -1.7 and 150-134/20= 0.8

Step 3: Lastly we find the z score of both results and subtract them and get our final answer.

(Z score of 0.8) .7881- .0446 ( Z score of -1.7) = .7435

]]>-Mr. Reitz

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probalility diamond 13/52 +probaility queen 4/52 minus 1/52 becuase out of the 13 diamonds is a queen of diamonds so we subtract one so as not to repeat a card

so 16/52 is the probability for diamond or queens, we can simplify so the answer is 4/13

]]>**Step 1:** First, we need to find our mean, standard deviation, and x. The mean and standard deviation are given to us in thew problem, but I found x because she needed to finish her project by 6 and she started at 5:30, so that would give her 30 minutes.

**Step 2: **Then, I plugged the values in the equation for Normal Distributions, which is: z = (x-mean)\(standard deviation), which is (30-25)/(9). That works ouot to be (5)/(9), which comes out to 0.555555556.

**Step 4:** Next, I rounded my answer. Rounding is very important for these type of problems. I rounded z=0.555555556 to z=0.56.

**Step 5: **After, I looked z=0.56 up in the table and the probability in the chart was .7123.

I hope you all are able to understand this problem. I did take a picture, but unfortunately, I’m having trouble uploading it. I’ll probably upload it in the near future once I get technical assistance. Study hard! Godd luck on the final!

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Problem 8:

The following is the probability distribution of Jeanette’s study hours for Mathematics in a given week.

X (hours) | 2 | 4 | 5 | 6 |

P(X) | 0.21 | 0.39 | ? | .15 |

a) Find the probability of *X=5*.

b) Find the mean of the probability distribution.

c) Find the standard deviation of the probability distribution.

Answers

a) The easiest way to find the probability of X=5 is adding all the other X’s and subtracting 1 from the answer. The result of the subtraction is the probability of X=5

Steps:

0.21+0.39+.15= .75 → 1-.75= .25 → P(5)= .25

Reason: The probability distribution p(X) should always equals 1 when you add them up.

b) Find the mean of the probability distribution:

Steps: By making a probability distribution chart: ↓

x |
P(x) |
xp(x) |
x² p(x) |

2 |
0.21 |
.42 |
.84 |

4 |
0.39 |
1.56 |
6.24 |

5 |
0.25 |
1.25 |
6.25 |

6 |
0.15 |
.9 |
5.4 |

∑(x)= 7 |
∑p(x)= 1 |
∑xp(x)=4.13 |
∑ x²p(x)= 18.73 |

µ= mean/average

∑xp(x)= µ µ= 4.13 µ²= 17.0569

To find the standard deviation, you have to use the formula: ∑ x²p(x) – µ² ← has to be square rooted

the answer is…1.29

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**A.** Construct a frequency distribution using 6 classes.

Subtract the highest value (276) by the lowest value (220) to get 56. Divide 56 by the number of classes (6) to obtain 9.3. Round 9.3 down to 9.

Select the lower limit of the first class. The lowest limit to 220

Add the class width to the lower limit of the first class to calculate the upper limit of the first class and the lower limit of the next class. Continue until all classes are completed. Add 9 to 220 to get the first class (220-229) and continue as follows:

(220-229)

(230-239)

(240-249)

(250-259)

(260 – 269)

(270-279)

Determine the frequencies for each class by counting the number of data values that fit for each class. The total frequency value should be equal to the total number of data values. Given the student scores:

(220-229) 3

(230-239) 4

(240-249) 5

(250-259) 5

(260 – 269) 2

(270-279) 1

**B.** Draw a histogram for the frequency distribution in part (a)

* X-Axis is for the frequency distribution class, and the Y-Axis is for the amount in those classes.

**C.** Draw a frequency polygon for the frequency distribution in part a

A Frequency polygon uses dots to connect the data instead of a straight line or bars

* X-Axis is for the frequency distribution class, and the Y-Axis is for the amount in those classes.

Good Luck,

Glen.

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Melissa Alteon

Question #16: A survey claims that a college graduate from Smith College can expect an average starting salary of $42,000. Fifteen Smith College graduates, drawn from normally distributed population, had an average starting salary of $40,800 with a standard deviation of $2,250. At the 1% level of significance, can we conclude that the average starting salary of the graduates is significantly less than $42,000?

Step 1: State the claim. Identify the null Hypothesis H0 and alternative hypothesis Ha.

Ho = µ = $42,000

Ha = µ ≠ $42,000

Step 2: Specify the level of significance . The level of significance gives the area of the rejection region(s)

α= .01, because 1% is equivalent to .01

Step 3: Describe the tails of the test. Sketch the rejection region(s)

The test is a two-tailed test since it is NOT equal to $42,000 that only means that it could be either GREATER than $42,000 or LESS than $42,000.

Step 4: Determine the degree of freedom d.f.= n-1

n=15, therefore n-1 would be 15-1 which equals 14

d.f. = 14

Step 5: Determine the critical value(s) using the t-distribution table.

This table is “Table 5- t-Distribution” and how you would determine the critical value is by reading the table left to right. We know that the d.f. = 14 and the test is a two-tailed test and that = .01 so following this method you descend from the value of alpha and you should have gotten

z= 2.977

Step 6: Find the t-value of the test static (from your sample).

z= (x-bar) – µ(x-bar)/ σ(x-bar)

µ (x-bar)= µ, µ= $42,000

σ (x-bar)= , s/ (square root of n)

s= $2250 and n= 15

s/(square root of n) = 580.95

x-bar = $40,80

z= ($40,800-$42,000)/580.95

z=-2.07

Step 7: Make a decision to reject or fail to reject the null hypothesis.

In this case the null hypothesis cannot be rejected because the t-value collected from the sample -2.07 does not lie outside the critical value instead it lies in between the critical values of -2.977 and +2.977. In other words it is not in the rejection area.

Step 8: Interpret the decision in the context of the original claim.

Therefore, A college graduate from Smith College can expect an average starting salary of $42,000 and we can conclude that the average starting salary of the graduates is less than $42,000.

**Step 4: Determine the degree of freedom d.f.= n-1**

n=15, therefore n-1 would be 15-1 which equals 14

d.f. = 14

**Step 5: Determine the critical value(s) using the t-distribution table.**

This table is “Table 5- t-Distribution” and how you would determine the critical value is by reading the table left to right. We know that the d.f. = 14 and the test is a two-tailed test and that = .01 so following this method you descend from the value of alpha and you should have gotten

**z= ** **2.977**

** **

**Step 6: Find the t-value of the test static (from your sample).**

= $42,000

, s= $2250 and n= 15

= 580.95

= $40,800

**Step 7: Make a decision to reject or fail to reject the null hypothesis.**

In this case the null hypothesis **cannot** be rejected because the t-value collected from the sample does not lie outside the critical value instead it lies in between the critical values of -2.977 and +2.977

**Step 8: Interpret the decision in the context of the original claim.**

Therefore, A college graduate from Smith College can expect an average starting salary of $42,000 and we can conclude that the avg starting salary of the graduates is less than $42,000.

Problem#12 on Exam#3 review sheet: There is an 80% chance that it will rain on any given day in seattle. What is the probability that, of 118 randomly selected days there will be rain on at most 90 of them?

Step 1: We figure out what n, p, q, and x first. n= 118, p= .8, q= .2, x= 1 to 90

Step 2: We figure out if both np and nq is greater than 5. np= 94.4, nq= 23.6 and both np and nq are greater than 5 so we can use the normal distribution.

Step 3: We have to figure out what the µ is and what σ is. µ= 94.4 and σ= 4.345112196

Step 4: We now convert the x to z. Because it is “at most” we add .5 to our original x before then apply that number which it is 90.5 to the formula. z=-.8975602526 and we round it to the nearest tenth place so z=-.9.

Step 5: We look up the z in the chart and the probability is 0.1841

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