Homework Assignment Week 3 Bridge Course

  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)

First, find the standard deviation for this set (the square root of 40,000). Standard Deviation = 200. Find the Z value for a steer at 979 lbs.

\[Z = \frac{x - \mu}{sd}\]

Z = (979 - 1300) / (200) = -1.605 Use the Z table to convert into percentage. 0.054247 = 5.42% However, the question asks for the probability that the weight of a randomly selected steer is greater than 979 lbs. So the answer is ** 1 - 0.054247 = 0.9457 or 94.57% **

print(round(pnorm(979, mean = 1300, sd = 200, lower.tail=FALSE),4))
## [1] 0.9458
  1. 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)

The calculations are similar to question 1. However, will apply the pnorm() function to obtain this answer.

print(round(pnorm(8340, mean = 11000, sd = sqrt(1960000), lower.tail=FALSE),4))
## [1] 0.9713
  1. 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)

To find this answer, we need to find the probability between 83 and 85 million dollars. To facilitate this, we need to find the lower.tail probability of 83 million and the higher.tail probability of 85 million. To find the answer, ** 1 - lower.tail probability of 83 million - higher.tail probability of 85 million **.

LessThan83 <- pnorm(83000000, mean=80000000, sd=3000000, lower.tail=TRUE)
MoreThan85 <- pnorm(85000000, mean=80000000, sd=3000000, lower.tail=FALSE)
Total <- LessThan83 + MoreThan85
Answer <- 1 - Total
print(round(Answer,4))
## [1] 0.1109
  1. 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.

We need to look for the minimum score required to be in the top 14%, or in other words, we need to find the score that matches with the 86% percentile.

print(round(qnorm(.86, mean=456, sd=123)),0)
## [1] 589
  1. 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.

Again, similar to question 4.

print(paste0("The lower limit is: ", round(qnorm(0.07, mean=6.13, sd=0.06), 2)))
## [1] "The lower limit is: 6.04"
print(paste0("The upper limit is: ", round(qnorm(0.93, mean=6.13, sd=0.06), 2)))
## [1] "The upper limit is: 6.22"
  1. 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.
HigherLimitOfC <- qnorm(0.55, mean=78.8, sd=9.8)
LowerLimitOfC <- qnorm(0.20, mean=78.8, sd=9.8)

print(paste0("The top limit of a C: ", round(HigherLimitOfC, 0)))
## [1] "The top limit of a C: 80"
print(paste0("The bottom limit of a C: ", round(LowerLimitOfC, 0)))
## [1] "The bottom limit of a C: 71"
  1. 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.
print(round(qnorm(0.55, mean=21.2, sd=5.4),1))
## [1] 21.9
  1. 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.)
Prob.Less.Than.11 <- dbinom(0, size=151, prob=0.09) + dbinom(1, size=151, prob=0.09) + dbinom(2, size=151, prob=0.09) + dbinom(3, size=151, prob=0.09) + dbinom(4, size=151, prob=0.09) + dbinom(5, size=151, prob=0.09) + dbinom(6, size=151, prob=0.09) + dbinom(7, size=151, prob=0.09) + dbinom(8, size=151, prob=0.09) + dbinom(9, size=151, prob=0.09) + dbinom(10, size=151, prob=0.09)
print(round(Prob.Less.Than.11, 4))
## [1] 0.192
  1. 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)

Because there is a sample of 147 tires, we must obtain the standard error of the mean. The equation for the standard error of the mean is: square root of the (variance/N) or SD / sqrt(N). Once, we get the standard error the mean, we can use this as our SD number and plug into the pnorm() function and obtain the Upper.Tail to find the probability of tire lifetime greater than 48.33 months.

Standard.Error.Mean <- 7/(sqrt(147))
print(round(pnorm(48.83, mean = 48, sd = Standard.Error.Mean, lower.tail=FALSE),4))
## [1] 0.0753
  1. 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)

Similar concept as question 9.

Standard.Error.Mean <- 10/(sqrt(68))
print(round(pnorm(93.54, mean = 91, sd = Standard.Error.Mean, lower.tail=FALSE),4))
## [1] 0.0181
  1. 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)

The mean of the sample proportion (p.hat) is p and the standard error of (p.hat) is s.e.(p.hat) = sqrt ( (p * (1-p) )/ n ). Therefore, the standard error of (p.hat) = sqrt (.07 * (1-.07)/540) = .01098.

So now we are attempting to find the probability of no shows between 4% to 10%, given the information we have.

Tenth.Percentile <- pnorm(.10, mean = .07, sd = .01098, lower.tail=TRUE)
Fourth.Percentile <- pnorm(.04, mean = .07, sd = .01098, lower.tail=TRUE)
Answer <- Tenth.Percentile - Fourth.Percentile
print(round(Answer,4))
## [1] 0.9937
  1. 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)

Similar to question 11.

S.E <- sqrt((.23 * (1 - .23))/602)
Twentyseventh.Percentile <- pnorm(.27, mean = .23, sd = S.E, lower.tail=TRUE)
Nineteenth.Percentile <- pnorm(.19, mean = .23, sd = S.E, lower.tail=TRUE)
Answer <- Twentyseventh.Percentile - Nineteenth.Percentile
print(round(Answer, 4))
## [1] 0.9803
  1. 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 (mean) = 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)

Calculate the margin of error. Z(alpha/2) * (sd of population) / sqrt(n). Z(.8/2) = Z(.4). Looking up on the z table, Z would equal 1.28 1.28 * (0.8) / sqrt(208) = .0710 Answer = mean +/- .0086 or 3.9 +/- .0710.

The upper limit is:

print(round(3.9 + .0710),1)
## [1] 4

The lower limit is:

print(round(3.9 - .0710,1))
## [1] 3.8
  1. 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 ® = 11 . Construct the 98% confidence interval for the mean per capita income. (Round your answers to 1 decimal place) (Sigma = 11)

Similar concept as question 13.

Z value that corresponds with Z(.98/2) = Z(.49) = 2.33

sd <- 11
mean <- 16.6
n <- 7472
Coefficient <- (2.33 * sd)/sqrt(n)
upper.limit <- mean + Coefficient
lower.limit <- mean - Coefficient
print(paste0("The upper limit is: ", round(upper.limit,1)))
## [1] "The upper limit is: 16.9"
print(paste0("The lower limit is: ", round(lower.limit,1)))
## [1] "The lower limit is: 16.3"
  1. 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.

Refer to the T statistics table. http://www.sjsu.edu/faculty/gerstman/StatPrimer/t-table.pdf

With 26 degrees of freedom, and with 1 tail, the value = 1.706 The best picture that responds to .05 of the area under the curve is image on the TOP RIGHT.

  1. 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)

vector.nums <- c(383.6, 347.1, 371.9, 347.6, 325.8, 337)
mean <- sum(vector.nums)/6
print(round(mean,2))
## [1] 352.17

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

S.D <- sd(vector.nums)
print(round(S.D,2))
## [1] 21.68

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

Typically, if the n number was high enough, we would refer to the Z table to obtain our values with alpha/2. However, given that the sample size is 6. Will need to refer to the T table with (n - 1) degrees of freedom, or in this case, it’s 5. Therefore, according to the table, the critical value is: 2.015

Critical.Value <- 2.015
print(Critical.Value)
## [1] 2.015

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

Upper.Limit <- mean + Critical.Value * S.D / sqrt(6)
Lower.Limit <- mean - Critical.Value * S.D / sqrt(6)
print(paste0("Upper Limit: ", round(Upper.Limit,2)))
## [1] "Upper Limit: 370"
print(paste0("Lower Limit: ", round(Lower.Limit,2)))
## [1] "Lower Limit: 334.34"
  1. 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)

Like the previous problem, whereas, we would have normally used a z table. Given that N = 16, will refer to T table with 15 degrees of freedom. 1.341

Critical.Value <- 1.341
print(round(Critical.Value,3))
## [1] 1.341

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

S.D <- 2.45
mean1 <- 46.4
Upper.Limit1 <- mean1 + Critical.Value * S.D / sqrt(16)
Lower.Limit1 <- mean1 - Critical.Value * S.D / sqrt(16)
print(paste0("Upper Limit: ", round(Upper.Limit1,1)))
## [1] "Upper Limit: 47.2"
print(paste0("Lower Limit: ", round(Lower.Limit1,1)))
## [1] "Lower Limit: 45.6"
  1. 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)

Arrange the formula Critical.Value * S.D / sqrt(n) = .13. –> (Critical.Value * S.D) / .13 ) ^2 = n

population.mean <- 8
population.sd <-1.9
#look for Z data for 99%.
z.99 <- 2.57
n <- round(((z.99 * population.sd)/0.13)^2)
print(n)
## [1] 1411
  1. 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)
bacteria.variance <- 3.61
bacteria.sd <- sqrt(3.61)
bacteria.mean <- 12.6
#look for Z data for 95%
z.95 <- 1.96
n <- round(((z.95 * bacteria.sd/0.19)^2))
print(round(n,0))
## [1] 384
  1. 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)

Below.8 <- 2089 - 1734
print(round(Below.8/2089,3))
## [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)

Using the margin of errors formula.

n = 2089
z.98 <- 2.33
Margin.Of.Error <- z.98 * sqrt((.17)*(1-.17)/n)
print(paste0("Upper limit: ", round(.17 + Margin.Of.Error,3)))
## [1] "Upper limit: 0.189"
print(paste0("Lower limit: ", round(.17 - Margin.Of.Error,3)))
## [1] "Lower limit: 0.151"
  1. 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)

Spilled <- 156/474
print(round(Spilled,3))
## [1] 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
z.95 <- 1.96
Margin.Of.Error <- z.95 * sqrt((.329)*(1-.329)/n)
print(paste0("Upper limit: ", round(.329 + Margin.Of.Error,3)))
## [1] "Upper limit: 0.371"
print(paste0("Lower limit: ", round(.329 - Margin.Of.Error,3)))
## [1] "Lower limit: 0.287"
  1. The cumulative distribution function of the random variable X is

F x to the power of (x) =1 - e to the power of (-a * x) , a greater than 0, x greater than 0

What is the probability density function? What is the expected value? What is the variance? Determine P(X<.5 | alpha =1).

The probability density function (PDF) P(x) of a continuous distribution is defined as the derivative of the (cumulative) distribution function. Therefore, the PDF function is: F’(x) = a * e^(-ax) .

?????Not 100% sure regarding expected value, variance or determinine the third question. I need to do some more reading and learning!

  1. The probability mass function for a particular random variable Y is

f(y)^y = ( e^(-b) * b^y) / y! .

Derive the mean and standard deviation.

This is the formula for the probability mass function for Poisson distribution. In this case, mean = (lambda or in this case, b) * t (where t is in time). Standard deviation is equal to sqrt(mean).