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Portage Learning Math 110 Final Exam
Question 1 You may find the following files helpful throughout the exam:
The following pie chart shows the percentages of total items sold in a month in a certain fast food restaurant.
A total of 4700 fast food items were sold during the
month.a.) How many were fish?
b.) How many were french fries?
Your Answer:
a. fish 4700(0.28)=
b. French fries 4700(0.4)=
a.) Fish : 4700(.28) = 1316
b.) French Fries: 4700(.40) = 1880
Question 2
Consider the following data: 430 389 414 401 466 421 399 387 450 407 392 410 440 417 471
Find the 40th percentile of this data.
There are a total of fifteen numbers, so n= 15. In order to find the percentiles, we must put the numbersin ascending order:
For the 40th percentile:
Therefore, the 40th percentile index for this data set is the 6th observation. In the list above, the 6thobservation is 407.
Question 3
In a tri-state conference, 60% attendees are from California, 25% from Oregon, and 15% from Washington. As it turns out 6 % of the attendees from California, 17% of the attendees from Oregon, and 12% of the attendees from Washington came to the conference by train. If an attendee is selected at random and found to have arrivedby train, what is the probability that the person is from Washington?
P(Train│C)=.06..
P(Train│O)=.17..P(Train│W)=.12..
P(C)=.60,P(O)=.25,P(W)=.15.
We want to find P(W│Train), so use:
Question 4
Find each of the following probabilities:
a. Find P(Z ≤ -0.87).
b. Find P(Z ≥ .93).
c. Find P(-.59 ≤ Z ≤ -.36).
a.
P(Z ≤ -0.87)= .19215.
C.
D.
E.
- List the values you would insert into that equation.
- State the final answer to the problem
We have a finite population, so we will use Case 2:
E.
The proportion of the sample that are defective is 15/75 = .2 so we set P=.2. As we mentioned previously, we estimate p by P. So, p=.2. A total of 75 monitors were tested, so n=75. Based on a confidence limit of 80 %, we find in table 6.1 that z=1.28. The total number of monitors is 450, so set N=450. Now, we can substitute all of these values into our equation:
So the proportion of the total that are defective is between .146 and .254.
Question 7 It is recommended that pregnant women over eighteen years old get 85 milligrams of vitamin C each day. The standard deviation of the population is estimated to be 9 milligrams per day. A doctor is concerned that her pregnant patients are not getting enough vitamin C. So, she collects data on 40 of her patients and finds that the mean vitamin intake of these 40 patients is 83 milligrams per day. Based on a level of significance of α = .015, test the hypothesis.
H 0 : μ=85 milligrams per day.
H 1 : μ<85 milligrams per day.
**This is a left-tailed test, so we must find a z that satisfies P(Z Passed Failed Drove 4 hours or less 250 80 Drove 8 hours or more 140 75
Is there is a relationship between hours of driving and alertness? (Do a test for independence.) Test at the 1 % level of significance.
H 0 : Driving hours and alertness are independent events.
H 1 : Driving hours and alertness are not independent events.
We have two rows and three columns, so # of Rows =2 and # of Columns=2. The degrees of freedom are given by:
DOF = (# of Rows-1)(# of Columns-1)=(2-1)(2-1)=1. Using this, along with .01 (for the 1% level of significance) we find in the chi-square table a critical value of 6.635.
This value is greater than the critical value of 6.635. So, we reject the null hypothesis.
