StudentGuiders

# ANLY 500 Homework-3

4.4. The Excel file *Atlanta Airline Data* provides arrival and taxi-in time statistics for one day at Atlanta Hartsfield International airport. Find the average and standard deviation of the difference between the scheduled and actual arrival times and the taxi-in time to the gate. Compute the *z*scores for each of these variables

Answer:
Average (Difference between scheduled and actual arrival time) is 4.91
Average (Taxi-in time) is 11.6
Standard Deviation (Difference between scheduled and actual arrival time) is 54.81
Standard Deviation(Taxi-in time) is 5.48
Z-scores:
**Z-score**
**Difference Z-Taxi-in**
**0.18409050.437956-0.1990513-1.0219-0.2537858-0.10949-0.1990513-0.474450.0016420.255474-0.4909688-1.02190.14760080.0729930.2753147-0.10949-0.1260719-0.29197-0.1625616-0.47445-0.2537858-0.83942-0.12607191.1678830.18409050.255474-0.2537858-0.65693-0.3997446-0.10949-0.107827-0.839420.3300493-0.83942-0.32676522.0802920.3665390.0729930.129356 1.715328 -0.4544791 -0.47445**
**-0.3267652 -0.83942**
And so on
4.22. The Excel file *Cell Phone Survey* reports opinions of a sample of consumers regarding the signal strength, value for the dollar, and customer service for their cell phone carriers. Use PivotTables to find the following:
1 the average signal strength by type of carrier
2 average value for the dollar by type of carrier and usage level
3 variance of perception of customer service by carrier and gender
Answer: 1. the average signal strength by type of carrier:
AT&T – 3.46
Sprint- 2.8
T-mobile – 3
Verizon – 3.8
Others – 2.67
**Row LabelsAverage of Signal strength**AT&T3.4Other2.666666667Sprint2.8T-mobile3Verizon3.8__(blank)5__** Grand Total 3.307692308**2. average value for the dollar by type of carrier and usage level
By Type of carrier and Usage Level

**Average of Value for the DollarColumn Labels**

__Row LabelsAverage__Hig__h__Lo**AT&T3.542.533.2Other2.833333333433Sprint4.66666666754.8T-mobile444Verizon2.666666667343.5**

__wVery highGrand Total____(blank)44__

**Grand Total3.39130434843.23.45454545 53.42307692 3**3. variance of perception of customer service by carrier and gender

**Var of Customer ServiceColumn LabelsRow LabelsFMGrand Total**AT&T0.6964285711.0955882350.973333333Other00.9523809520.777777778Sprint00.3333333330.2T-mobile#DIV/0!#DIV/0!0.5Verizon0.81.31.333333333(blank)#DIV/0!#DIV/0!

**Grand Total0.7614379080.9705882350.926093514**Verizon has highest signal strength, and sprint has the lowest signal strength but valued highly by customers. Verizon has high variance and Sprint has low variance 4.31. The Excel file

*Auto Survey*contains a sample of data about vehicles owned, whether they were purchased new or used, and other types of data. Use the

*Descriptive Statistics*tool to summarize the numerical data, find the correlations among each of the numerical variables, and construct PivotTables to find the average miles/gallon for each type of vehicle, and also the average miles/gallon and average age for each type of new and used vehicle. Summarize the observations that you can make from these results. Answer:

*Vehicle Age*
* *
*Mileage*
* *
*MPG*
* *
8.20652 100914. 29.0782
Mean 2 Mean 5 Mean 6
1.01892 13212.0 1.84601
Standard Error 3 Standard Error 1 Standard Error 9
Median 9 Median 105628 Median 28.7
Mode 7 Mode #N/A Mode 20
Standard 4.88658 Standard 63362.5 Standard 8.85319
Deviation 4 Deviation 6 Deviation 4
23.8787 4.01E+0 78.3790
Sample Variance 1 Sample Variance 9 Sample Variance 5
Kurtosis -1.28727 Kurtosis -1.27265 Kurtosis -0.42041
Skewness -0.28886 Skewness -0.0667 Skewness 0.1023
Range 14.75 Range 201834 Range 33.7
Minimum 0.25 Minimum 3454 Minimum 12
Maximum 15 Maximum 205288 Maximum 45.7
Sum 188.75 Sum 2321034 Sum 668.8
__Count 23 Count 23 Count 23__
**Row LabelsAverage of **
**MPG**Large SUV18.1Mid-size 33.52
Minivan 16
Small 33.37272727
Small SUV 21.96666667
**Grand Total29.07826087** **Row LabelsAverage of Vehicle AgeAverage of **
**MPG** Large SUV10.518.1 Mid-size5.7533.52 Minivan11.516 Small7.72727272733.37272727 __Small SUV10.3333333321.96666667__ **Grand Total8.20652173929.07826087** **Average of **
**MPGColumn Labels**__ Row LabelsNewUsedGrand Total__Large SUV2115.218.1 Mid-size37.327.8533.52 Minivan201216 Small32.0666666734.9433.3727272
7
21.9666666

**Grand Total30.8557**Small SUV 24.9

**Average of Vehicle**

**AgeColumn Labels**

__Large SUV71410.5 Mid-size0.58333333313.55.75 Minivan91411.5 Small5.66666666710.27.72727272 7 10.3333333__

**Row LabelsNewUsedGrand Total****Grand Total4.64583333399**Small SUV 4 4.38. A Midwest pharmaceutical company manufactures individual syringes with a self-contained, single dose of an injectable drug. In the manufacturing process, sterile liquid drug is poured into glass syringes and sealed with a rubber stopper. The remaining stage involves insertion of the cartridge into plastic syringes and the electrical “tacking” of the containment cap at a precisely determined length of the syringe. A cap that is tacked at a shorterthan-desired length (less than 4.920 inches) leads to pressure on the cartridge stopper and, hence, partial or complete activation of the syringe. Such syringes must then be scrapped. If the cap is tacked at a longer-thandesired length (4.980 inches or longer), the tacking is incomplete or inadequate, which can lead to cap loss and a potential cartridge loss in shipment and handling. Such syringes can be reworked manually to attach the cap at a lower position. However, this process requires a 100% inspection of the tacked syringes and results in increased cost for the items. This final production step seemed to be producing more and more scrap and reworked syringes over successive weeks. The Excel file

*Syringe Samples*provides samples taken every 15 minutes from the manufacturing process. Develop control limits using the data and use statistical thinking ideas to draw conclusions.

**Answer**: Population mean = 4.9581 5.4. A market research company surveyed consumers to determine their ranked preferences of energy drinks among the brands Monster, Red Bull, and Rockstar. 1 What are the outcomes of this experiment for one respondent? 2 What is the probability that one respondent will rank Red Bull first? 3 What is the probability that two respondents will both rank Red Bull first? Answer: a. Monster Red Bull Rockstar Monster Rockstar Red Bull Red Bull Rockstar Red Bull Red Bull Red Bull Rockstar Rockstar Monster Red Bull Rockstar Red Bull Monster b. 3 possibilities, so probability is 1/3 c. Based on b, one respondent’s probability is 1/3. So, for 2 respondent’s is (1/3)*(1/3) = 1/9 5. 9. An airline tracks data on its flight arrivals. Over the past 6 months, 50 flights on one route arrived early, 150 arrived on time, 25 were late, and 45 were canceled. 1. What is the probability that a flight is early? On time? Late? Canceled? 2. Are these outcomes mutually exclusive? 3. What is the probability that a flight is either early or on time? Answer: a. What is the probability that a flight is early? On time? Late? Canceled? Flight is early: 50/270 = 5/27 Flight is on time: 150/270 = 5/9 Flight is late: 25/270 =5/54 Flight is cancelled: 45/270 = 9/54 = 1/6 b. Yes. c. probability that a flight is either early or on time. 5/27 + 5/9 = 20/27 5.14. In an example in Chapter 3, we developed the following cross-tabulation of sales transaction data:

__East564298North434285South623799West10090190Total2612114721. Find the marginal probabilities that a sale originated in each of the four regions and the marginal probability of each type of sale (book or DVD). 2. Find the conditional probabilities of selling a book given that the customer resides in each region. Answer 1. A. East: P(east) = 98/472 = 0.2076 B. North: P(North) = 85/472 = 0.18 C. South: P(South) = 99/472 = 0.2097 D. West: P(West) = 190/472 = 0.402 P(Book) = 261/472 = 0.55 P(DVD) = 211/472 = 0.44 Answer 2. Find the conditional probabilities of selling a book given that the customer resides in each region. P(Book | East) = P(Book and East)/P(East) = (56/472) / 0.2076 = 0.57 P(Book|West) = 0.526 P(Book|North) = 0.505 P(Book|South) = 0.626 5.22. The weekly demand of a slow-moving product has the following probability mass function:__

**Region Book DVD Total**

**Demand,**

**x**

**Probability,**

**f**

**(**

**x**__0 0.2 1 0.4 2 0.3 3 0.1 4 or more 0 Find the expected value, variance, and standard deviation of weekly demand.__

**)****Answer**: Expected value= 0.4*1 + 0.3*2 + 0.1*3 – 1.3 Variance = 0.8, Std dev= 0.9 5.33. Verify that the function corresponding to the following figure is a valid probability density function. Then find the following probabilities: 1

*P*(

*x*< 8) 2

*P*(

*x*> 7) 3

*P*(6 <

*x*< 10) 4

*P*(8 <

*x*< 11) Answer: Condition to be satisfied is area under graph =1 and f(x) > 0 for x. Since area under graph = 0.3+.3+.1+.1+.1+.1=1, therefore it is valid probability density function. Note(I used ref: https://newonlinecourses.science.psu.edu/stat414/node/97/) 1.

*P*(

*x*< 8) = 0.3+0.3=0.6 2.

*P*(

*x*> 7) = 0.3+0.1+0.1+0.1+0.1=0.7 3.

*P*(6 <

*x*< 10) = 0.3+0.3+0.1+0.1=0.8 4.

*P*(8 <

*x*< 11) = 0.1+0.1+0.1 = 0.3 5.36. In determining automobile-mileage ratings, it was found that the mpg (

*X*) for a certain model is normally distributed, with a mean of 33 mpg and a standard deviation of 1.7 mpg. Find the following: 1

*P*(

*X*< 30) 2

*P*(28 <

*X*< 32) 3

*P*(

*X*> 35) 4

*P*(

*X*> 31) 5 The mileage rating that the upper 5% of cars achieve. Answer: a. P(X<30): Take X=30, Z = X-mean / sd = (30-33)/1.7 = -1.76 Given Z score of -1.76, P is 0.039 from the table. b. P(28<X<32) = P(28-33 / 1.7 < Z < 32-33/1.7) = P(-2.94 < Z < -0.588) P(-2.94 < Z < -0.588) = P(Z< -0.588) - P(Z< -2.94) P(Z< -0.588) = 1- P(Z< 0.588), based on P ( Z< −a)=1−P ( Z< a ) P(Z< 0.588) = 0.7224 using standard normal table P(Z< -0.588) = 1−0.7224 = 0.2776 P(Z< -2.94) = 1- P(Z< 2.94) = 1−0.9984=0.0016 Therefore, P(-2.94 < Z < -0.588) = 0.276 Note: Reference: https://www.mathportal.org/calculators/statisticscalculator/normal-distribution-calculator.php c.

*P(X > 35)*Take X=35, Z = X-mean / sd = (35-33)/1.7 = 1.176,

*From the table,*Given Z score of 1.176, P is 0.879 d.

*P(X > 31), Take X = 31,*Z = X-mean / sd = (31-33)/1.7 = -1.176

*From the table,*Given Z score of -1.176, P is 0.121 e.

*The mileage rating that the upper 5% of cars achieve. P(X<35.8) = 0.95*5.39. A supplier contract calls for a key dimension of a part to be between 1.96 and 2.04 centimeters. The supplier has determined that the standard deviation of its process, which is normally distributed, is 0.04 centimeter. 1 If the actual mean of the process is 1.98, what fraction of parts will meet specifications? 2 If the mean is adjusted to 2.00, what fraction of parts will meet specifications? 3 How small must the standard deviation be to ensure that no more than 2% of parts are nonconforming, assuming the mean is 2.00?

*Answer: Reference for standard normal distributed table: http://sphweb.bumc.bu.edu/otlt/MPHModules/BS/BS704_Probability/BS704_Probability9.html*a.

*P(X<1.96) =*Take X=1.96, Z = X-mean / sd = (1.96-1.98)/0.04 = -0.5,

*From the table,*Given Z score of -0.5, P is 0.308

*P(X<2.04) =*Take X=

*2.04*, Z = X-mean / sd = (

*2.04*-1.98)/0.04 = 1.5,

*From the table,*Given Z score of 1.5, P is 0.933 fraction of parts will meet specifications between 1.96 and 2.04 centimeters = 0.9330.308=0.625 -> 62.5% b. Mean=2,

*P(X<1.96) =*Take X=1.96, Z = X-mean / sd = (1.96-2)/0.04 = -1,

*From the table,*Given Z score of -1, P is 0.158

*P(X<2.04) =*Take X=

*2.04*, Z = X-mean / sd = (

*2.04*-2)/0.04 = 1,

*From the table,*Given Z score of 1, P is .84 fraction of parts will meet specifications between 1.96 and 2.04 centimeters = 0.84-0.158= 0.682 > 68.2% c. Mean=2,let’s take S.D. to be 0.017 cm,

*P(X<1.96) =*

*Take X=1.96, Z = X-mean / sd = (1.96-2)/0.017 = -2.35*

*From the table, Given Z score of -2.35, P is 0.0093*

*P(X<2.04) =*

*Take X=2.04, Z = X-mean / sd = (2.04-2)/0.017 , From the table, Given Z score , P is 0.9907*

*fraction of parts will meet specifications between 1.96 and 2.04 centimeters = 0.9907-*

*0.0093= 0.9814 > 98.14%*

*So by taking*

**sd = 0.0172**, we get fraction of parts will meet specifications between 1.96 and*2.04 centimeters = 98%*5.42. The actual delivery time from Giodanni’s Pizza is exponentially distributed with a mean of 20 minutes. 1 What is the probability that the delivery time will exceed 30 minutes? 2 What proportion of deliveries will be completed within 20 minutes?

*Answer:*probability that the delivery time will exceed 30 minutes: P(X<30), X = 30, mean =20

*So, it is 0.*77 proportion of deliveries will be completed within 20 minutes: P(X<20), with mean =20

*So, it is 0.*62