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)
Here,
Mean (M) = 1300 lbs
Variance (V) = 40,000
X = 979 lbs
Standard Deviation (SD) = ???Variance
= ???40000
In R,
X <- 979
M <- 1300
V <- 40000
SD <- sqrt(S)
round(1 - pnorm(x , m, sd),4)
= 0.9458
Therefore, the probability that the randomly selected steer is greater than 979 lbs. is 94.58%.
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)
Here,
Mean (M) = 11000 hrs
Variance (V) = 1,960,000 lbs
X = 8340 hrs
Standard Deviation (SD) = ???Variance
= ???1960000
In R,
X <- 8340
M <- 11000
V <- 1960000
SD <- sqrt(V)
round(1 - pnorm(X , M, SD),4)
= 0.9713
Therefore, the probability that the randomly selected steer is greater than 8340 lbs. is 97.13%.
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)
Here,
Mean (M) = 80000000
Standard Deviation (SD) = 3000000
X1 = 83000000
X2 = 85000000
Variance (V) = SD2
In R,
X1 <- 83000000
X2 <- 85000000
M <- 80000000
SD <- 3000000
# for lower number 83 million dollar
Px1 <- round(pnorm(X1 , M, SD),4)
Px1
# Px1 = 0.8413
# for higher number 85 million dollar
Px2 <- round(pnorm(X2 , M, SD),4)
Px2
# Px2 = 0.9522
# Probability
Px = Px2 - Px1
Px
# Px = 0.1109
Therefore, the probability that a randomly selected firm will earn between 83 million to 85 million dollars is 11.09%.
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.
Here,
Mean (M) = 456
Standard Deviation (SD) = 123
Suppose the Total score = 100
According to the question,
Top scorers selected for job = 14%
= 0.14
So, Bottom scorers = 1 - 0.14
= 0.86
In R
#top 14#
X1 <- 0.14
#bottom = 1 - top
X2 <- 1 - X1
M <- 456
SD <- 123
round(qnorm(X2, M, SD),0)
= 589
Therefore, the minimum score required for the job is 589.
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.
Here,
Mean (M) = 6.13 cm
Standard Deviation (SD) = 0.06 cm
Top 7% is actually 93%
Bottom 7% is actually 7%
In R,
#top 7% is 93%#
X1 <- 0.93
#bottom 7% is 7%
X2 <- 0.07
M <- 6.13
SD <- 0.06
# for the top 7%
round(qnorm(X1, M, SD),4)
# top 7% = 6.2185
# for the bottom 7%
round(qnorm(X2, M, SD),4)
# bottom 7% = 6.0415
Therefore, the two lengths that separate the top 7% and bottom 7% are 6.2185 cm and 6.0415 cm respectively.
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.
Here,
Mean (M) = 78.8
Standard Deviation (SD) = 9.8
For C, Top 45% = 0.55
Bottom 20% = 0.20
In R,
#top 45%
X1 <- 0.55
#bottom 20%
X2 <- 0.20
M <- 78.8
SD <- 9.8
# for the top 45%
round(qnorm(X1, M, SD),0)
# 80
# for the bottom 20%
round(qnorm(X2, M, SD),0)
# 71
Therefore, the values to get C are between 71 and 80.
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.
Here,
Mean (M) = 21.2
Standard Deviation (SD) = 5.4
For Top 45% = 0.55
Bottom 20% = 0.20
In R,
#top 45% for acceptance
X1 <- 0.55
M <- 21.2
SD <- 5.4
# for the top 45%
round(qnorm(X1, M, SD),0)
# minimum score is 22.
Therefore, the minimum score required for the admission is 22.
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.)
Here,
N = 151
P = 0.09
Since, we know that the probability of students not graduating is less than 11.
X = 10
In R,
X <- 10
N <- 151
P <- 0.09
# for the distribution we use pbinom
round(pbinom(X, N, P), 4)
# 0.192
Therefore, the probability using the normal distribution will be 19.2%
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)
Here,
Mean (X1) = 48 months
Mean(X2) = 48.83 months
Standard Deviation (SD1) = 7
Sample Size (N) = 147
In R,
X1 <- 48
X2 <- 48.83
N <- 147
SD1 <- 7
SD2 <- SD1/sqrt(N)
#Since, months are greater, there is higher chance the probability is also going be more
round(1 - pnorm(X2, X1, SD2),4)
# 0.0753
Therefore, the probability that mean of the sample would be greater than 48.83 is 7.53%.
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)
Here,
Mean (X1) = 91 months
Standard Deviation (SD) = 10
Sample of Computers (N) = 68
Mean(X2) = 93.54
In R,
X1 <- 91
X2 <- 93.54
N <- 68
SD1 <- 10
SD2 <- SD1/sqrt(N)
#Since, months are greater, there is higher chance the probability is also going be more
round(1 - pnorm(X2, X1, SD2),4)
# 0.0181
Therefore, the probability that mean of the sample of 68 computers is 1.81%.
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)
Here,
Using Fisher Transformation for r:
z: = 1/2 lna???(1+r)/(1-r)???
Standard Error = 1/???(N-3)
In R,
z <- function(r)
return(.5*(log((1+r)/(1-r))))
SE <- function(n)
return(1/sqrt(n-3))
# to find area between 4% and 10%
M <- z(.07)
SD <- SE(540)
P4 <- pnorm(z(.04), M, SD)
P10 <- pnorm(z(.10), M, SD)
round(P10 - P4, 4)
#0.5153
Therefore, the probability of no show could differ by 51.53%.
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)
Here,
M <- z(.23)
sem <- sem(602)
# to find area greater than 19% and less than 27%
P19 <- pnorm(z(.19), M, sem)
P27 <- pnorm(z(.27), M, sem, lower.tail=FALSE)
round(P19 + P27, 4)
# 0.301
Therefore, the probability that the proportion of defective bottles would differ is 30.1%.
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 ?????? = 3.9. Assume å = 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)
Lower Bound:
Upper Bound:
Here,
By using Confidence Interval formula with unknown Standard Deviation,
CI = µ ± tn-1??/(???N)
µ = 3.9
?? = 0.8
N = 208
In R,
M <- 3.9
SD <- 0.8
N <- 208
CI <- 0.8
P <- ( 1 - CI ) / 2
# By using formula we can calculate our t value.
t <- qt(P, N-1)
CI1 <- round(M + t * SD / sqrt(N),1)
CI2 <- round(M - t * SD / sqrt(N),1)
CI1
# 3.8
CI2
# 4
Therefore, the confidence interval for the consumption of beef per week by age over 48 is between 3.8 to 4 lbs.
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 ?????????? = 16.6. Assume å = 11 . Construct the 98% confidence interval for the mean per capita income. (Round your answers to 1 decimal place)
Lower Bound:
Upper Bound:
Here,
µ = 16.6
?? = 11
N = 7472
In R,
#14
M <- 16.6
SD <- 11
N <- 7472
CI <- 0.98
P <- (1 - CI ) / 2
# By using formula we can calculate our t value.
t <- qt(P, N-1)
CI1 <- round(M + t * SD / sqrt(N),1)
CI2 <- round(M - t * SD / sqrt(N),1)
CI1
# 16.3
CI2
# 16.9
Therefore, the 98% confidence interval for the mean per capita income is between 16.3 to 16.9 thousand dollars.
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
Here,
N = 26
Area under the curve p = 0.05
In R,
#15
N <- 26
P <- 0.05
# By using formula
t <- round(qt(P, N-1),4)
t
# -1.7081
Therefore, the value of t = -1.7081 or -t = 1.7081.
Step 2. Write your answer below.
The picture that represents the answer is on the top right.
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)
Here,
According to the question,
S = {383.6, 347.1, 371.9, 347.6, 325.8, 337}
In R,
s <- c(383.6, 347.1, 371.9, 347.6, 325.8, 337)
m <- round(mean(s),2)
m
# 352.17
Therefore, the sample mean for the given data is 352.17.
Step 2. Calculate the sample standard deviation for the given sample data. (Round answer to 2 decimal places)
Here,
Standard Deviation
In R,
SD <- round(sd(s, na.rm = FALSE),2)
SD
# sd = 21.68
Therefore, the standard deviation is 21.68.
Step 3. Find the critical value that should be used in constructing the confidence interval. (Round answer to 3 decimal places)
Here,
s <- c(383.6, 347.1, 371.9, 347.6, 325.8, 337)
N <- length(s)
CI <- 0.90
P <- ( 1 - CI ) / 2
# By using formula we can calculate our t value.
t <- round(qt(P, N-1), 3)
t
# t = -2.015
Therefore, the critical value t = -2.015 or -t = 2.015.
Step 4. Construct the 90% confidence interval. (Round answer to 2 decimal places)
Lower Bound
Upper Bound
Here,
CI1 <- round(M + t * SD / sqrt(N),2)
CI2 <- round(M - t * SD / sqrt(N),2)
CI1
# CI1 = 334.34
CI2
# CI2 = 370
Therefore, the 90% confidence interval is 334.34 to 370.70.
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)
Here,
µ = 46.4
?? = 2.45
N = 16
In R,
M <- 46.4
SD <- 2.45
N <- 16
CI <- 0.80
P <- ( 1 - CI ) / 2
# By using formula we can calculate our t value.
t <- qt(P, N-1)
round(t,3)
# t = -1.341
Therefore, the critical value is 1.341.
Step 2. Construct the 80% confidence interval. (Round answer to 1 decimal place)
Lower Bound:
Upper Bound:
Here,
CI1 <- round(M + t * SD / sqrt(N),1)
CI2 <- round(M - t * SD / sqrt(N),1)
CI1
# CI1 = 45.6
CI2
# CI2 = 47.2
Therefore, the 80% confidence interval is between 45.6 and 47.2.
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)
Here,
µ = 8
?? = 1.9
Confidence interval = 99%
Standard Error SE = 0.13
In R,
M <- 8
SD <- 1.9
# Standard Error
SE <- 0.13
#Confidence interval 99% has z value of 2.576
z <- 2.576
#calculate for N
N <- round((z * SD / SE)^2, 0)
N
# N = 1417
Therefore, the sample size is 1417.
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)
Here,
µ = 12.6
?? = 3.61
Confidence interval = 95%
Standard Error SE = 0.19
In R,
M <- 12.6
SD <- sqrt(3.61)
# Standard Error
SE <- 0.19
#Confidence interval 95% has z value of 1.96
z <- 1.96
N <- round((z * SD / SE)^2,0)
N
# N = 384
Therefore, the sample size is 384.
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)
Here,
Sample Size (N) = 2089
Students above eighth grade level (x) = 1734
In R,
N <- 2089
x <- 1734
P <- round(1 - (x/N),3)
P
# P = 0.17
Therefore, the proportion of tenth graders reading at or below the eight grade level is 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)
Lower Bound
Upper Bound
Here,
N = 2089
Xabove = 1734
Xbelow = 2089 - 1734
= 355
CI = 98% reading below eight grade level
In R,
N <-2089
Xabove <- 1734
Xbelow <- N - Xabove
#calculate standard error
SEbelow <- sqrt((P*(1-P))/Xbelow)
# z value for 98% confidence level is 2.326
z <- 2.326
# calculating the intervals,
CI1 <- round(P - z * SEbelow, 3)
CI2 <- round(P + z * SEbelow, 3)
CI1
# CI1 = 0.124
CI2
#CI2 = 0.216
Therefore, 98% confidence interval for tenth grade readers is between 0.124 to 0.216.
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)
Lower Bound:
Upper Bound:
Here,
Sample size (N) = 474
Spills (s) = 156
In R,
N <- 474
s <- 156
P <- round(s/N, 3)
P
# P = 0.329
Therefore, the proportion of spill on oil tankers 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)
Here,
Sample size (N) = 474
Spills (s) = 156
Confidence Interval (CI) = 95%
In R,
N <-474
s<- 156
sbelow <- N - s
# calculate Standard Error
SEbb <- sqrt((P*(1-P))/sbelow)
# z value for 95% confidence level is 1.96
z <- 1.96
#calculating the intervals,
CI1 <- round(P - z * SEbb, 3)
CI2 <- round(P + z * SEbb, 3)
CI1
# CI1 = 0.277
CI2
# CI2 = 0.381
Therefore, the 95% confidence interval is between 0.277 to 0.381.