Test Name: test2

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)

# mean  = μ = 1300
# variance = σ^2 = 40000
#standar deviation is the sqare root of variance
#x>= 949
x = 949
m<-mean(1300)
m

## [1] 1300

v<-variance <- 40000
v

## [1] 40000

stddev<-sqrt(variance)
stddev

## [1] 200

#standard deviation = 200
round(pnorm(x , m, stddev, lower = FALSE),4)

## [1] 0.9604

Answer: The probability that the weight of a randomly selected steer is greater than 979 lbs is 0.9604

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)

#Variance= σ^2 = = 1960000
#mean == μ = 11000
#x = 8340
x <- 8340
x

## [1] 8340

m<-mean(11000)
m

## [1] 11000

v<-variance <- 1960000
stddev<-sqrt(variance)
stddev

## [1] 1400

#standard deviation = 1400
round(pnorm(x, m, stddev, lower = FALSE),4)

## [1] 0.9713

The probability that the life span of the monitor will be more than 8340 hours is 0.9713.

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)

 #x between 83 and 85 million
 m<-mean(80)
m

## [1] 80

stddev<-3
round(pnorm(85, m, stddev, lower = TRUE) - pnorm(83, m, stddev, lower = TRUE),4)

## [1] 0.1109

The probability that a randomly selected firm will earn between 83 and 85 million dollars is 0.1109

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.

#Variance= σ^2 = =
#mean == μ = 456
#x = 
x <- 0.14
x

## [1] 0.14

m<-mean(456)
m

## [1] 456

stddev<-123
stddev

## [1] 123

#standard deviation = 1400
round(qnorm(x, m, stddev, lower = FALSE),0)

## [1] 589

The minimum score required for the job offer is 589.

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.

#x =  .07  for bottom and 1-.07 = 0.93 for top
x1 <-0.07
x2<-0.93
m<-mean(6.13)
m

## [1] 6.13

stddev<-.06
stddev

## [1] 0.06

round(qnorm(x1, m, stddev),3)

## [1] 6.041

round(qnorm(x2, m, stddev),3)

## [1] 6.219

The top 7% rejected is greater than 6.219 cm. The bottom 7% rejected is less than 6.041 cm.

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.

#x =  limit for B score is <= 55% and greater then 20
x1 <-.55
x1

## [1] 0.55

x2 <-.201
x2

## [1] 0.201

m<-mean(78.8)
m

## [1] 78.8

stddev<-9.8
stddev

## [1] 9.8

round(qnorm(x1, m, stddev),0)

## [1] 80

round(qnorm(x2,m, stddev),0)

## [1] 71

The numerical limits to a grade of C lies greater then equal to 71 and less than or equal to 80 (71-80).

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.

#x =  minimum score 1-.45 = .55
x1 <-.55
x1

## [1] 0.55

x2 <-
x2
m<-mean(21.2)
m

## [1] 21.2

stddev<-5.4
stddev

## [1] 5.4

round(qnorm(x1, m, stddev),1)

## [1] 21.9

The minimum score required is 21.9.

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.)

#Number in trail = 151
# probability  = .09
N <- 151
p <- .09
m <- N * p
s <- N * p * (1 - p)
stddev <- sqrt(s)
p <- round(pnorm(11, m, stddev, lower=TRUE),4)
p

## [1] 0.2307

Answer: Approximate graduation on time based on normal distribution is 0.2307.

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)? # https://www.researchgate.net/post/Should_I_use_the_standard_deviation_or_the_standard_error_of_the_mean

#https://www.researchgate.net/post/Which_of_the_following_measures_is_better_to_show_the_differences

x <-48.83
n <- 147
m <- 48
stddev <- 7
stderr_mean <- stddev / sqrt(n) 
p <- round(pnorm(x, m, stddev, lower=FALSE),4)
p

## [1] 0.4528

p2 <- round(pnorm(x, m, stderr_mean, lower=FALSE),4)
p2

## [1] 0.0753

#probability of .4528 seema too high almost 50/50. Think anwer is more reasonable at 0.0753 per the above article.

Answer: The probability of the mean being greater then 48.83 months is 0.0753.

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)

x <-93.54
 N <- 68
m <- 91
stddev <- 10
stderr_mean <- stddev / sqrt(N) 
p1 <- round(pnorm(x, m, stddev, lower=FALSE),4)
p1

## [1] 0.3997

p2 <- round(pnorm(x, m, stderr_mean, lower=FALSE),4)
p2

## [1] 0.0181

Answer: The probability that the mean of a sample of 68 computers would be greater than 93.54 months is 0.0181

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)

sqrt(π(1−π)/N) π = m

x1 <-.10
x2<-.04
 N <- 540
m <-0.07
π <-0.07
stderr_mean <-sqrt((π*(1-π)/N))
stderr_mean

## [1] 0.01097978

p1 <- pnorm(x1, m, stderr_mean, lower=TRUE)
p1

## [1] 0.9968553

p2 <- pnorm(x2, m, stderr_mean, lower=TRUE)
p2

## [1] 0.003144737

p3<-round(p1-p2 ,4)
p3

## [1] 0.9937

Answer: 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% is 0.9937

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)

x1 <-.19
x2<-.27
 N <- 602
m <-0.23
π <-0.23
stderr_mean <-sqrt((π*(1-π)/N))
stderr_mean

## [1] 0.01715185

p1 <- pnorm(x1, m, stderr_mean, lower=TRUE)
p1

## [1] 0.009847463

p2 <- 1-pnorm(x2, m, stderr_mean, lower=TRUE)
p2

## [1] 0.009847463

p3<-round(p1+p2 ,4)
p3

## [1] 0.0197

Answer: The probability that the proportion of defective bottles in a sample of 602 bottles would differ from the population proportion by greater than 4% is 0.0197.

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 R =.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)

80% confidence level = 1.282 z level

x=3.9
n <- 208
σ <- .8
z <-1.282 
σ_m <- σ / sqrt(n)
lower <- round(x - (z * σ_m),1)
lower

## [1] 3.8

upper <- round(x + (z * σ_m),1)
upper

## [1] 4

Answer: Upper Limit is 3.8 Lower Limit is 4

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 r=11  .  Construct the 98% confidence interval for the mean per capita income. (Round your answers to 1 decimal place)

#98% confidence level = 2.326 z level

x=16.6
n <- 7472
x <- 3.9
σ <- 11
z <-2.326 
σ_m <- σ / sqrt(n)
lower <- round(x - (z * σ_m),1)
lower

## [1] 3.6

upper <- round(x + (z * σ_m),1)
upper

## [1] 4.2

Answer: Lower bound = 3.6 Upper bound = 4.2

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.

(Pictures won’t copy in r) Step 1. Choose the picture which best describes the problem.

Answer: The picture to the top right

Step 2. Write your answer below. # http://www.sjsu.edu/faculty/gerstman/StatPrimer/t-table.pdf df = 26 and one tail for value .05 is 1.706 from the table

Answer: 1.706

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)

x <- c(383.6, 347.1, 371.9, 347.6, 325.8, 337)
m <- round(mean(x),2)
m

## [1] 352.17

Answer: 352.17 picocuries per liter

Step 2. Calculate the sample standard deviation for the given sample data. (Round answer to 2 decimal places)

x <- c(383.6, 347.1, 371.9, 347.6, 325.8, 337)
stddev <- round(sd(x),2)
stddev

## [1] 21.68

Answer: 21.68 is the standard deviation for sample

Step 3. Find the critical value that should be used in constructing the confidence interval. (Round answer to 3 decimal places)

Answer: T-table -t95 df 6

The critical value is 1.943

Step 4. Construct the 90% confidence interval. (Round answer to 2 decimal places)

x <- c(383.6, 347.1, 371.9, 347.6, 325.8, 337)
t <- 1.943
m <- 352.17
stddev <- 21.68
s2 <- stddev / sqrt(m)
lower <- x - (t * s2)
lower

## [1] 381.3553 344.8553 369.6553 345.3553 323.5553 334.7553

upper <- x + (t * s2)
upper

## [1] 385.8447 349.3447 374.1447 349.8447 328.0447 339.2447

Answer: Lower is 334.76 and the upper is 339.25

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)

http://www.sjsu.edu/faculty/gerstman/StatPrimer/t-table.pdf

t= 80 two tails df:16-1 = 15

Answer: 0.866

Step 2. Construct the 80% confidence interval. (Round answer to 1 decimal place)

n <- 16
x <- 46.4
stddev <- 2.45
t <- 0.866
s2 <- stddev / sqrt(n)
lower <- x - (t * s2)
lower

## [1] 45.86957

upper <- x + (t * s2)
upper

## [1] 46.93042

Answer: Lower is 45.9 and upper is 46.9

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)

https://people.richland.edu/james/lecture/m170/ch08-int.html z = 2.576 for 99% confidnece margin of error: z*(stddev/sqrt(n))

n = (z*stddev/s2)^2

z<-2.576
stddev = 1.9
s2 = .13
n<- (z*stddev/s2)^2
n

## [1] 1417.465

Answer: The answer is number of population needs 1418

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)

 v <- 3.61
s <- sqrt(v)
z <- 1.96
s2 <- 0.19
n <- ( (z * s) / s2)^2
n

## [1] 384.16

Answer: Sample size is 385

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)

p <- 1 - (1734 / 2089)
round(p,3)

## [1] 0.17

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)

n <- 2089
s <- sqrt( (p * (1 - p)) / n )
z <- 2.326
lower <-round(p - (z * s) - (0.5 / n),3)
upper <- round(p + (z * s) + (0.5 / n),3)
lower

## [1] 0.151

upper

## [1] 0.189

Answer: The lower is 0.151 and upper is 0.189

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)

p <- 156 / 474
round(p,3)

## [1] 0.329

Answer: Probabliliyt is 0.329

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)

n <- 474
s2 <- sqrt( (p * (1 - p)) / n )
z <- 1.960
lower <- p - (z * s2) - (0.5 / n)
round(lower,3)

## [1] 0.286

upper <- p + (z * s2) + (0.5 / n)
round(upper,2)

## [1] 0.37

Answer: Lower is 0.286 and upper is 0.37