# 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)
stndev=sqrt(40000)
mn=1300
round(pnorm(979,mn,stndev,lower.tail=FALSE),4)
## [1] 0.9458
# 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)
stndev=sqrt(1960000)
mn=11000
round(pnorm(8340,mn,stndev,lower.tail=FALSE),4)
## [1] 0.9713
# 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)
stndev=sqrt(3)
mn=80
round(pnorm(85,mn,stndev)- pnorm(83,mn,stndev),4)
## [1] 0.0397
# 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.
stndev=123
mn=456
round(qnorm(.86,mn,stndev),0)
## [1] 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.
stndev=0.06
mn=6.13
round(qnorm(.93,mn,stndev),2)
## [1] 6.22
round(qnorm(.07,mn,stndev),2)
## [1] 6.04
# 6. An English professor assigns letter grades on a test according to the following scheme.
stndev=9.8
mn=78.8
round(qnorm(.55,mn,stndev))
## [1] 80
round(qnorm(.20,mn,stndev))
## [1] 71
# 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.
stndev=5.4
mn=21.2
round(qnorm(.55,mn,stndev),0)
## [1] 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.)
round(pbinom(10,151,.09),4)
## [1] 0.192
# 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)
# fnc to get stndrd error of mean
getSEM=function(std, n){
return(std/sqrt(n))
}
pmn = 48
sd = 7
n = 147
sem=getSEM(sd, n)
round(pnorm(48.83, pmn, sem, lower.tail = FALSE),4)
## [1] 0.0753
#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)
pmn = 91
sd = 10
n = 68
sem=getSEM(sd, n)
round(pnorm(93.54, pmn, sem, lower.tail = FALSE),4)
## [1] 0.0181
#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)
prsonZ=function(r){
return(.5*(log((1+r)/(1-r))))
}
prSEM=function(n){
return(1/sqrt(n-3))
}
#Find area between 4% and 10$
mn=prsonZ(.07)
sem=prSEM(540)
p4=pnorm(prsonZ(.04),mn,sem)
p10=pnorm(prsonZ(.10),mn,sem)
round(p10 - p4 , 4)
## [1] 0.5153
#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)
mn=prsonZ(.23)
sem=prSEM(602)
#Find area <19% and area >27%
p19=pnorm(prsonZ(.19),mn,sem)
p27=pnorm(prsonZ(.27),mn,sem,lower.tail = FALSE)
round(p19 + p27, 4)
## [1] 0.301
#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 x̅ = 3.9. Assume sigma = 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)
smplmn = 3.9
stndev=0.8
n=208
t=abs(qt(.2/2,n-1))
se=sd/sqrt(n)
lower = round(smplmn - t*se, 1)
upper = round(smplmn + t*se, 1)
lower ; upper
## [1] 3
## [1] 4.8
#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 x̅ = 16.6. Assume sigma = 11 . Construct the 98% confidence interval for the mean per capita income. (Round your answers to 1 decimal place)
smplmn = 16.6
stndev= 11
n=7472
t=abs(qt(.02/2,n-1))
se=sd/sqrt(n)
lower = round(smplmn - t*se, 1)
upper = round(smplmn + t*se, 1)
lower ; upper
## [1] 16.3
## [1] 16.9
#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 : Upper Right
#Step 2
d = 26
p = .05
round(qt(p,d),4)
## [1] -1.7056
# 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
s <- c(383.6, 347.1, 371.9, 347.6, 325.8, 337)
n <- length(s)
smn <- round(sum(s)/n, 2)
smn
## [1] 352.17
#Step 2
sd <- round(sqrt(var(s)), 2)
sd
## [1] 21.68
#Step 3
t <- round(abs(qt(.10/2, n-1)), 4)
t
## [1] 2.015
#Step 4
se <- sd/sqrt(n)
lower <- round(smn - t*se, 2)
upper <- round(smn + t*se, 2)
lower; upper
## [1] 334.34
## [1] 370
#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.
n <- 16
smn <- 46.4
sd <- 2.45
# Step 1
t <- round(abs(qt(.2/2, n-1)), 3)
t
## [1] 1.341
#Step 2
se <- sd/sqrt(n)
lower <- round(smn - t*se, 1)
upper <- round(smn + t*se, 1)
lower; upper
## [1] 45.6
## [1] 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)
pmn = 8
sd = 1.9
z = 2.575 #two sided 99 confd level
n= round(((z*sd)/0.13)^2)
n
## [1] 1416
#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)
pmn= 12.6
var = 3.61
sd = sqrt(var)
#95% Confidence.
z <- 1.96
n <- round(((z*sd)/0.19)^2)
n
## [1] 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
#sample prob <= 8
n <- 2089 #CLT Applies
p <- 1 - 1734/n
round(p, 3)
## [1] 0.17
#Step 2
z <- 2.33
se <- sqrt((p*(1-p)) / n)
cc <- 0.5/n
lower <- round(p - z * se - cc, 3)
upper <- round(p + z * se + cc, 3)
lower; upper
## [1] 0.151
## [1] 0.189
#21. An environmentalist wants to find out the fraction of oil tankers that have spills each month.
#Step 1:
n= 474
phat = 156/n
round(phat, 3)
## [1] 0.329
#Step 2
z=1.96
se = sqrt((phat*(1-phat))/n)
cc = 0.5/n
lower=round(phat - z * se - cc, 3)
upper=round(phat + z * se + cc, 3)
lower; upper
## [1] 0.286
## [1] 0.372
#22. The Cumulative Distribution Function
#Fx^(x) = 1-e^-ax, a>0, x>0
#Exponential distribution function, CDF is given
#Step 1: Probability density fn -> take the 1st derivation of CDF
#Ans = ae^(-ax)
#Step 2 Expected Value
#Ans = 1/a
#Step 3 Variance
#Ans = 1/a^2
#Step 4
#P(X<.5 | alpha =1)
x= 0.5
a= 1
pdf = a * exp(-a * x)
pdf
## [1] 0.6065307
#**************************************
#23 . The probability mass function for a particular random variable Y is
#fy^(y)= e^-b*b^y/y!
#Step 1 Mean
#Ans = b
#Step 2 Variance
#Ans = b