Data-Science-Math-Week-3-Assignment
Murat Anil Akyildirim
7/30/2019
Question 1
The weights of steers in a herd are distributed normally. The variance is 40,000 and the mean steer weight is 1300 lbs. Find the probability that the weight of a randomly selected steer is greater than 979 lbs. (Round your answer to 4 decimal places)
Answer 1
This is a normal distribution.
Variance = 40000
Mean = 1300 lbs
Standard Deviation is the square root of Mean.
m1 <- 1300
v1 <- 40000
sd1 <- sqrt(v1)
sd1## [1] 200Standard variance is 200.
We are being asked probability of the weight greater than 979.
## Probability of having a value *lower* than 979 in a normal distribution with a mean of 1300 and sd of 200.
p1 <- pnorm(q = 979, mean = 1300, sd = 200, lower.tail = FALSE)
p1## [1] 0.9457531We need to round it to 4 as part of the question.
round(p1,4)## [1] 0.9458The answer of question 1 is 0.9458.
Question 2.
SVGA monitors manufactured by TSI Electronics have life spans that have a normal distribution with a variance of 1,960,000 and a mean life span of 11,000 hours. If a SVGA monitor is selected at random, find the probability that the life span of the monitor will be more than 8340 hours. (Round your answer to 4 decimal places)
Answer 2.
This is a normal distribution as outlined in the question.
Variance = 1960000
Mean = 11000
Standard deviation is square root of the variance.
v2 <- 1960000
m2 <- 11000
sd2 <- sqrt(v2)
sd2## [1] 1400Standard Deviation is 1400.
The question is the random variable that has a life span of the monitor over 8340 hours.
## Probability of having a value *lower* than 8340 in a normal distribution with a mean of 11000 and sd of 1400.
p2 <- pnorm(q = 8340, mean = 11000, sd = 1400, lower.tail = FALSE)
p2## [1] 0.9712834We need to round the number to 4 decimal as part of the question.
round(p2,4)## [1] 0.9713The answer of question 2 is 0.9713.
Question 3
Suppose the mean income of firms in the industry for a year is 80 million dollars with a standard deviation of 3 million dollars. If incomes for the industry are distributed normally, what is the probability that a randomly selected firm will earn between 83 and 85 million dollars? (Round your answer to 4 decimal places)
Answer 3
This is a normal distribution as it is informed by the question.
Standard Deviation is 3 million
Mean is 80 million
The question is probability of firm earning being between 83 and 85 million.
We can find the probability of earning being below 83 first:
## Probability of having a value *lower* than 83 in a normal distribution with a mean of 80 and sd of 3.
p3 <- pnorm(q = 83, mean = 80, sd = 3, lower.tail = FALSE)
p3## [1] 0.1586553We can round to 4 decimals.
round(p3, 4)## [1] 0.1587We can find the probability of earnings being below 85 million.
## Probability of having a value *lower* than 85 in a normal distribution with a mean of 80 and sd of 3.
p4 <- pnorm(q = 85, mean = 80, sd = 3, lower.tail = FALSE)
p4## [1] 0.04779035We need to round to 4 decimal.
round(p4, 4)## [1] 0.0478The difference between the probability being below 85 and probability between 83 will give us the probability that is P(83<x<85).
p5 <- 0.1587 - 0.0478
p5 ## [1] 0.1109The answer of question 3 is 0.1109.
Question 4
Suppose GRE Verbal scores are normally distributed with a mean of 456 and a standard deviation of 123. A university plans to offer tutoring jobs to students whose scores are in the top 14%. What is the minimum score required for the job offer? Round your answer to the nearest whole number, if necessary.
Answer 4
This is a normal distribution as it is outlined in the question.
Mean is 456
Standard Deviation is 123
Top 14 % will get a scholarship. 14% is 0.14
## X-axis value *below* which the probability (AUC) is 14%
q1 <- qnorm(p = 0.14, mean = 456, sd = 123, lower.tail = FALSE)
q1## [1] 588.8793We need to round to the nearest whole number which is 588. The answer of question 4 is 588.
Question 5
The lengths of nails produced in a factory are normally distributed with a mean of 6.13 centimeters and a standard deviation of 0.06 centimeters. Find the two lengths that separate the top 7% and the bottom 7%. These lengths could serve as limits used to identify which nails should be rejected. Round your answer to the nearest hundredth, if necessary.
Answer 5
This is a normal distribution as stated in the question.
Mean is 6.13 cm
Standard deviation is 0.06 cm
The question is to find the length that separates 7% bottom and 7% top.
## X-axis value *below* which the probability (AUC) is 7% . This will give me the bottom number that will separate the 7%
q2 <- qnorm(p = 0.07, mean = 6.13, sd = 0.06, lower.tail = FALSE)
q2## [1] 6.2185476.22 cm is the number that seperates the bottom 7%
curve(dnorm(x, mean = 6.13, sd = 0.06), xlim = c(5, 7))
abline(h = 0)
sequence <- seq(0, 98.36897, 0.1)
polygon(x = c(sequence,98.36897,0),
y = c(dnorm(c(sequence),6.13,0.06),0,0),
col = "grey")
## X-axis value *below* which the probability (AUC) is 93% .
q3 <- qnorm(p = 0.93, mean = 6.13, sd = 0.06, lower.tail = FALSE)
q3## [1] 6.0414536.0 is the number that seperates the top 7%. The answer of question 5 is 6.22 cm and 6 cm
Question 6
An English professor assigns letter grades on a test according to the following scheme. A: Top 13% of scores B: Scores below the top 13% and above the bottom 55% C: Scores below the top 45% and above the bottom 20% D: Scores below the top 80% and above the bottom 9% F: Bottom 9% of scores Scores on the test are normally distributed with a mean of 78.8 and a standard deviation of 9.8. Find the numerical limits for a C grade. Round your answers to the nearest whole number, if necessary.
Answer 6
This is a normal distribution as outlined in the question.
Mean = 78.8
Standard Deviation is 9.8
The question is to find the numerical limit for C grade which is top 45% and above the bottom 20%
so top 45 % is 55% bottom
so we are looking for x value probability between 20% and 55%.
## X-axis value *below* which the probability (AUC) is 20% .
q4 <- qnorm(p = 0.20, mean = 78.8, sd = 9.8, lower.tail = TRUE)
q4## [1] 70.55211## X-axis value *below* which the probability (AUC) is 55% .
q5 <- qnorm(p = 0.55, mean = 78.8, sd = 9.8, lower.tail = TRUE)
q5## [1] 80.03148C grade will between 70 and 80. The answer of question 6 is bottom limit is 70 and top limit is 80.
Question 7
Suppose ACT Composite scores are normally distributed with a mean of 21.2 and a standard deviation of 5.4. A university plans to admit students whose scores are in the top 45%. What is the minimum score required for admission? Round your answer to the nearest tenth, if necessary.
Answer 7
This is a normal distribution as outlined in the question.
Top 45% scores (0.45) will be accepted.
Standard deviation is 5.4
Mean is 21.2
The question is what is the minimum score required.
## X-axis value *below* which the probability (AUC) is 45% .
q6 <- qnorm(p = 0.45, mean = 21.2, sd = 5.4, lower.tail = TRUE)
q6## [1] 20.52143The answer of question 7 is 21.
Question 8
Consider the probability that less than 11 out of 151 students will not graduate on time. Assume the probability that a given student will not graduate on time is 9%. Approximate the probability using the normal distribution. (Round your answer to 4 decimal places.)
Answer 8
This is a normal distribution.
Probability for less than 11 out of 151 is 11/151. (less than 11 so lower than or equal to 10)
Probability for graduation is 9 %. 0.09
Total students (sample size) is 151
## Probability of having a value *lower* than or equal to 10 in a binomial distribution with 151 trials with success probability of 0.09. P[X <= x]. Include the "equal to" value.
p6 <- pbinom(q = 10, size = 151, prob = 0.09, lower.tail = TRUE)
p6## [1] 0.191967We need to round to 4 decimal places.
round(p6, 4)## [1] 0.192The answer of question 8 is 0.192
Question 9
The mean lifetime of a tire is 48 months with a standard deviation of 7. If 147 tires are sampled, what is the probability that the mean of the sample would be greater than 48.83 months? (Round your answer to 4 decimal places)
Answer 9
Sample size is 147.
Standard deviation is 7
Mean is 48
Probability that the mean of the sample is higher than 48.83 months.
We need to find out the standard deviation for sample size 147.
standard deviation is square root of the variance
standard deviation for the sample size will be. standard deviation / square root of the sample size
sd3 <- 7/sqrt(147)
sd3## [1] 0.5773503## Probability of having a value *lower* than 48.3 in a normal distribution with a mean of 48 and sd of 0.577.
p7 <- pnorm(q = 48.83, mean = 48, sd = 0.577, lower.tail = FALSE)
p7## [1] 0.07514968The answer of question 9 is 0.075
Question 10
The quality control manager at a computer manufacturing company believes that the mean life of a computer is 91 months, with a standard deviation of 10. If he is correct, what is the probability that the mean of a sample of 68 computers would be greater than 93.54 months? (Round your answer to 4 decimal places)
Answer 10
Standard Deviation is 10.
Mean is 91.
Sample size is 68 computers.
Probability of the mean of greater than 93.54 months.
Same way as question 9 , we can find the standard deviation based on 68 computers.
sd4 <- 10/sqrt(68)
sd4## [1] 1.212678The standard deviation is 11.03537 for sample size 68.
## Probability of having a value *lower* than 93.54 in a normal distribution with a mean of 91 and sd of 1.212.
p8 <- pnorm(q = 93.54, mean = 91, sd = 1.212, lower.tail = FALSE)
p8## [1] 0.01805398**The answer of question 10 is 0.018*
Question 11
A director of reservations believes that 7% of the ticketed passengers are no-shows. If the director is right, what is the probability that the proportion of no-shows in a sample of 540 ticketed passengers would differ from the population proportion by less than 3%? (Round your answer to 4 decimal places)
Answer 11
0.07 ticketed passengers are no show is directors assumption.
sample size is 540.
The question is, probability of the proportion of no shows different than the population proportion by less than 3%.
so I think what is being asked is P(7-3%<x<7+3%) = P(4%<x<10%)
Calculating standard deviation:
sd10 <- sqrt(.07*(1-.07)/540)
sd10## [1] 0.01097978## Probability of having a value *lower* than 0.04 in a normal distribution with a mean of and sd of 0.0109.
pnorm(q = 0.04, mean = 0.07, sd = 0.0109, lower.tail = FALSE)## [1] 0.997041The answer of question 11 is 0.0071
## Probability of having a value *lower* than 0.10 in a normal distribution with a mean of and sd of 0.0109.
pnorm(q = 0.10, mean = 0.07, sd = 0.0109, lower.tail = FALSE)## [1] 0.002958972# difference between each probability
0.997041-0.00295## [1] 0.994091The answer of question 11 is 0.994
Question 12
A bottle maker believes that 23% of his bottles are defective. If the bottle maker is accurate, what is the probability that the proportion of defective bottles in a sample of 602 bottles would differ from the population proportion by greater than 4%? (Round your answer to 4 decimal places)
Answer 12
0.23 defective bottles
Sample size is 602
greater than 0.04
P(0.19<x<0.27) -same approach as question 11
Calculating standard deviation:
sd11 <- sqrt(0.23*(1-0.23)/602)
sd11## [1] 0.01715185## Probability of having a value *lower* than 0.19 in a normal distribution with a mean of and sd of 0.01715.
pnorm(q = 0.19, mean = 0.23, sd = 0.01715, lower.tail = FALSE)## [1] 0.9901592## Probability of having a value *lower* than 0.27 in a normal distribution with a mean of and sd of 0.01715.
pnorm(q = 0.27, mean = 0.23, sd = 0.01715, lower.tail = FALSE)## [1] 0.00984084# difference between each probability
0.9901592-0.00984084## [1] 0.9803184The answer of question 12 is 0.980
Question 13
A research company desires to know the mean consumption of beef per week among males over age 48. Suppose a sample of size 208 is drawn with x ̅ = 3.9. Assume ® = 0.8 . Construct the 80% confidence interval for the mean number of lb. of beef per week among males over 48. (Round your answers to 1 decimal place)
Mean consumption of males over 48
sample size is 208
mean is 3.9
standard deviation is 0.8
80 % confidence interval for the mean of lb of beef per week among males over 48.
Confidence interval formula is.
We are looking for 80% confidence level.
error <-qt(0.80, df=208-1)*0.8/sqrt(208)
error## [1] 0.04678126ci1 <- 3.9-error
ci2 <- 3.9+error
ci1## [1] 3.853219ci2## [1] 3.946781The answer of question 13 is 3.9 and 4
Question 14
An economist wants to estimate the mean per capita income (in thousands of dollars) in a major city in California. Suppose a sample of size 7472 is drawn with x ̅ = 16.6. Assume ® = 11 . Construct the 98% confidence interval for the mean per capita income. (Round your answers to 1 decimal place)
Answer 14
Same approach as question 13.
Sample size is 7472
Mean is 16.6
Standard Deviation is 11
Confidence interval we are looking to construct is 98%
error2 <-qt(0.98, df=7472-1)*11/sqrt(7472)
error2## [1] 0.2613951ci3 <- 16.6 + error2
ci4 <- 16.6 - error2
ci3## [1] 16.8614ci4## [1] 16.3386The answer of question 14 is 16.8 and 16.3
Question 15
Find the value of t such that 0.05 of the area under the curve is to the left of t. Assume the degrees of freedom equals 26.
Step 1. Choose the picture which best describes the problem
Step 2. Write your answer below.
Answer 15
Question 16
The following measurements ( in picocuries per liter ) were recorded by a set of helium gas detectors installed in a laboratory facility:
383.6, 347.1, 371.9, 347.6, 325.8, 337 Using these measurements, construct a 90% confidence interval for the mean level of helium gas present in the facility. Assume the population is normally distributed.
Step 1. Calculate the sample mean for the given sample data. (Round answer to 2 decimal places)
Step 2. Calculate the sample standard deviation for the given sample data. (Round answer to 2 decimal places)
Step 3. Find the critical value that should be used in constructing the confidence interval. (Round answer to 3 decimal places)
Step 4. Construct the 90% confidence interval. (Round answer to 2 decimal places)
Answer 16
Let’s create the data set first.
helium <- c(383.6, 347.1, 371.9, 347.6, 325.8, 337)
helium## [1] 383.6 347.1 371.9 347.6 325.8 337.0Let’s get the mean as part of the step 1 question
m1 <- mean(helium)
m1## [1] 352.1667Let’s round to 2 decimals.
round(m1,2)## [1] 352.17The answer of question 16, step 1 is 352.17
sd4 <- sd(helium)
sd4## [1] 21.67585Let’s round to 2 decimal.
round(sd4,2)## [1] 21.68The answer of question 16 , step 2 is 21.68
critical_1 <-qt(0.1/2, 6-1)
critical_1## [1] -2.015048The answer of question 16 , step 3 is -2.015
In step 4, we need to figure out the error and + and minues from the mean for confidence interval.
error3 <-qt(0.90, df=6-1)*21.68/sqrt(6)
error3## [1] 13.06279ci5 <- 352.17 + error3
ci6 <- 352.17 - error3
ci5## [1] 365.2328ci6## [1] 339.1072The answer of question 16 step 4 is 365.23 and 339.10
Question 17
A random sample of 16 fields of spring wheat has a mean yield of 46.4 bushels per acre and standard deviation of 2.45 bushels per acre. Determine the 80% confidence interval for the true mean yield. Assume the population is normally distributed.
Step 1. Find the critical value that should be used in constructing the confidence interval. (Round answer to 3 decimal places)
Step 2. Construct the 80% confidence interval. (Round answer to 1 decimal place)
Answer 17
Mean is 46.4
standard deviation is 2.45
sample size is 16
critical_2 <- qt(0.20/2, 16-1 )
critical_2## [1] -1.340606The answer of question 17 step 1 is -1.341
error4 <-qt(0.80, df=16-1)*2.45/sqrt(16)
error4## [1] 0.530575ci6 <- 46.4 + error4
ci7 <- 46.4 - error4
ci6## [1] 46.93058ci7## [1] 45.86942round(ci6, 1)## [1] 46.9round(ci7, 1)## [1] 45.9The answer of question 17 step 2 is 45.9 and 46.9
Question 18
A toy manufacturer wants to know how many new toys children buy each year. She thinks the mean is 8 toys per year. Assume a previous study found the standard deviation to be 1.9. How large of a sample would be required in order to estimate the mean number of toys bought per child at the 99% confidence level with an error of at most 0.13 toys? (Round your answer up to the next integer)
Answer 18
Mean is 8 toys per year.
Standard deviation is 1.9.
Standard Error is 0.13.
The have all the variable information except sample size, which is being asked here and the z value. The confidence level is 99%. z score is 2.58
Sample size is (z value * standard deviation/standard error)^2
sample_size <- (2.58*1.9/0.13)^2
sample_size## [1] 1421.87round(sample_size, 0)## [1] 1422The answer of question 18 is 1422
Question 19
A research scientist wants to know how many times per hour a certain strand of bacteria reproduces. He believes that the mean is 12.6. Assume the variance is known to be 3.61. How large of a sample would be required in order to estimate the mean number of reproductions per hour at the 95% confidence level with an error of at most 0.19 reproductions? (Round your answer up to the next integer)
Answer 19
Mean is 12.6
Variance is 3.61
Sample size is being asked
Confidence level is 95%
Standard error is 0.19.
We can have the similar approach as question 18.
Based on the table the z score for 95% is 1.96.
## we need to find the standard devation based on our formula.
sd5 <- sqrt(3.61)
sample_size_2 <- (1.96*sd5/0.19)^2
sample_size_2## [1] 384.16round(sample_size_2, 0)## [1] 384The answer of question 19 is 384
Question 20
The state education commission wants to estimate the fraction of tenth grade students that have reading skills at or below the eighth grade level.
Step 1. Suppose a sample of 2089 tenth graders is drawn. Of the students sampled, 1734 read above the eighth grade level. Using the data, estimate the proportion of tenth graders reading at or below the eighth grade level. (Write your answer as a fraction or a decimal number rounded to 3 decimal places)
Step 2. Suppose a sample of 2089 tenth graders is drawn. Of the students sampled, 1734 read above the eighth grade level. Using the data, construct the 98% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level. (Round your answers to 3 decimal places)
Answer 20
Sample size is 2089
1734 read the above the eight grade level.
Proportion of tenth graders reading at or below the eight level is being asked
## Proportion of having a value *higher* than or equal to 1734.
proportion <- 1-(1734/2089)
proportion## [1] 0.1699378The answer of question 20 step 1 is 0.17
Total numner of students 2089. Total number of students that reads like an 8th grader is 1734. So 2089-1734 are the amount of students that does not read like an 8th grader.
rest_students <- 2089-1734
rest_students## [1] 355The same approach as the earlier confidence interval questions. Find the error and add/minues from the proportion.
based on the chart above z is 2.33 for 98% confidence interval.
error10 <-qt(0.17, df=2089-1)*2.33/sqrt(2089)
error10## [1] -0.048653ci10 <- proportion + error10
ci11 <- proportion - error10
ci10## [1] 0.1212848ci11## [1] 0.2185908The answer of question 20 is 0.12 and 0.21
Question 21
An environmentalist wants to find out the fraction of oil tankers that have spills each month.
Step 1. Suppose a sample of 474 tankers is drawn. Of these ships, 156 had spills. Using the data, estimate the proportion of oil tankers that had spills. (Write your answer as a fraction or a decimal number rounded to 3 decimal places)
Step 2. Suppose a sample of 474 tankers is drawn. Of these ships, 156 had spills. Using the data, construct the 95% confidence interval for the population proportion of oil tankers that have spills each month. (Round your answers to 3 decimal places)
Answer 21
Sample size is 474
156 had spills and
474-156= 318 does not have spills.
The same approach as question 20.
## Proportion of having a value *higher* than or equal to 156.
proportion_2 <-156/474
proportion_2## [1] 0.3291139The answer of question 21 and step 1 is 0.329
Based on the chart 1.96 z score is for 95% confidence interval.
error11 <-qt(0.33, df=474-1)*1.96/sqrt(474)
error11## [1] -0.03962852ci12 <- proportion + error11
ci13 <- proportion - error11
ci12## [1] 0.1303092ci13## [1] 0.2095663The answer of question 21 step 2 is, 0.13 and 0.20