Home
Subjects
Explanations
Create
Study sets, textbooks, questions
Log in
Sign up
Upgrade to remove ads
Only $35.99/year
Final Exam Study Tool - Chapter 7
STUDY
Flashcards
Learn
Write
Spell
Test
PLAY
Match
Gravity
Isabelle, Lindsey, Justin, Erica
Terms in this set (20)
7.2
The mean is μ= 12,485 and the standard deviation is σ= 21,973. Find the probability that a random sample size of n=10 will have a mean number greater than 17,000
μ= 12,485
σ= 21,973
the population is skewed right
n≥30?
10≥30? no. we do not know if the sample distribution is approximately normal so we cannot find the probability.
7.2
The mean is = 12,485 and the standard deviation is σ= 21,973. Find the probability that a random sample size of n=36 will have a mean number greater than 17,000
μ= 12,485
σ= 21,973
the population is skewed right
n≥30?
36≥30? yes. the sampling distribution is approximately normal by the central limits theorem (CLT)
p(x̄>17,000)= normalcdf (upper bound, lower bound, μx̄, σx̄)
Mx̄= μ= 12,485
σx̄= σ/√n = 21,973/√36= 3662.17
normalcdf( 17,000, 1E99, 12,485, 3662.17) = 0.109
7.2
determine if the given set of data is approximately normal by creating a stat plot
x̄: 5, 6, 7, 5, 4, 3, 5, 6, 8, 3
data: L1
stat plot: on
zoom: 9
graph: linear so the data is approximately normal
7.2
If n>30 (n= sample size) then the data is approximately normal. complete the chart below to determine if the data is approximately normal
POP. SHAPE | SAMPLE SIZE | SAMPLING DIST. SHAPE
skewed left | 5
skewed right | 35
normal | 5
skewed left | 35
skewed right | 5
normal | 35
POP. SHAPE | SAMPLE SIZE | SAMPLING DIST. SHAPE
skewed left | 5 | skewed left
skewed right | 35 | approx. normal
normal | 5 | approx. normal
skewed left | 35 | approx. normal
skewed right | 5 | skewed right
normal | 35 | approx. normal
7.1
A retirement home is trying to figure out the probability of their residents being between the ages of 65 and 80 where the mean age is 70 with a standard deviation of 4.5.
μ= 70
σ= 4.5
normalcdf(lower bound, upper bound, mean, standard deviation)
normalcdf( 65, 80, 70, 4.5)
Answer: The probability is 85.361%
(7.1)
Commute times to colleges: We are interesting in how long it takes the 6 graduates in the top 1% of 2021 to commute to their colleges. The time (in hours) is given in the table. Since these 6 graduates are all of the top percent, we can consider them a population.
Brendan 10
Lily 24
Ben 36
Jacob 14
Emma 5
Gracie 1
a) Calculate the population mean
b) take a sample of Lily, Ben, Emma, and Gracie. Find the sample mean and calculate the sampling error.
a) 90/6 = 15
μ = 15
b) 24+36+5+1 = 66 /4 = 16.5
x̅ = 16.5
| x̅-μ | = | 16.5-15 | = 1.5
(7.1)
Scientists are interesting in finding the mean age of women who have recently given birth. In a sample size of 20 with a population standard deviation of 5 as well as a population mean of 5, find the sample mean.
μ = 25
μx̅ = μ
thus μx̅ = 25
(7.1)
Scientists are interesting in finding the mean age of women who have recently given birth. In a sample size of 20 with a population standard deviation of 5 as well as a population mean of 5, find the sample standard deviation.
σx̅ = (σ/√4)
σx̅ = 6/√20
σx̅ = 1.118
(7.2)
A bottling company uses a filling machine to fill plastic bottles with what is supposed to be 300ml of soda. However, the contents vary according to a normal distribution with mean μ=298ml with a standard deviation of σ=3ml. What is the probability that a randomly selected bottle contains less than 295 ml?
Approximately Normal-Stated
NormalCDF(-1E99, 295, 298, 3)
=.1586 or 15.86%
(7.1)
Suppose the scores for the ACT in 2021 are normally distributed with a mean of 25 and a standard deviation of 5. A sample of 25 students are selected. Find the 5th percentile of the sample
Inversenorm(left, μx̅, σx̅)
μx̅ = 25
σx̅ = (5/√25) = 1
Inversenorm (.05, 25, 1)
= 23.355
(7.1)
The service lifetime of a set of brake pads is 55,000 miles, with a standard deviation of 4,500 miles. If the National Highway Traffic Safety Administration randomly selects 8 cars for inspection, what is the distribution of the sample mean lifetime of the brake pads in the 8 cars?
μ=55,000 σ=4,500 n=8
μx̅=55,000 σx̅=4,500/sqrt(8)=1590.99
μx̅=55,000 and σx̅=1590.99
(7.1)
If the sample mean lifetime of a set of brake pads contains the distribution μ=55,000 miles and σ=4500 miles, and a random sample of 8 cars yields the results: μx̅=51,800 , σx̅=1591 what is the probability that a randomly chosen set would have a mean lifetime of less than 51,800 miles?
μx̅=51,800 and P(μx̅<51,800)
NormalCDF(-1E99, 51,800, 55,000, 1591)
P(μx̅<51,800)=0.0221
(7.1)
In a sample mean chosen at random, what is the probability of a value being chosen between 35 and 47? μx̅= 37 , σx̅ = 5
normalcdf (lower, upper, μx̅, σx̅)
normalcdf (35,47,37,5)
= .633 or 63.3%
(7.1)
The heights of a population of students have a mean of 68 inches with a standard deviation of 3 inches. If the sample size (n) is 1000, find μx̅ and σx̅.
μx̅=68
σx̅=σ/sqrt(n)
σx̅=3/31.62
σx̅=0.09 inches
(7.1)
The heights of a population of students have a mean of 68 inches with a standard deviation of 3 inches. A sample size (n) yields the following result μx̅=68. If the sample size (n) were lowered to 100, what would μx̅ equal?
μx̅=68
Changing the sample size would not change the value of μx̅.
(7.3)
25% of girls in a graduating class of 100 have blonde hair. Let P denote the population of girls with blond hair. find p̂, μp̂, and σp̂
p̂ = x/n = success over total
μp̂ = p̂
σp̂ = √((p(1-p)/n)
p̂ = 25/100 = .25
μp̂ = .25
σp̂ = √((.25(1-.25)/100) = .0433
(7.3)
Suppose that dogs are favored by 63% of families over cats in a district sample size of 200. Use p̂ to estimate p. What is the probability that p̂ >.5? Make sure to check for normal.
A) normal : np ≥ 10 n(1-p) ≥ 10
.63(200) = 126 ≥ 10 200(1-.63) = 74 ≥ 10
B) μp̂ = .63
σp̂ = √((.63(1-.63)/200) = .0341
normal cdf(area, 1E99, μp̂ , σp̂)
normalcdf( .5, 1E99, .63, .0341) = .9999
(7.1)
Time to commute to work: We are interesting in the time it takes a family to get to their respective jobs. The time (in minutes) is given in the table. Since the 5 members in the family all work, we can consider them a population.
Alyssa 45
Brent 30
Hayden 15
Camden 20
Caleb 10
a) calculate the population mean
b) take a sample of Brent, Hayden, and Caleb. Find the sample mean and calculate the sampling error.
a) 120/5 = 24
μ = 24
b) 30+15+10 = 55 / 3 = 18.333
x̅ = 18.333
| x̅-μ | = | 18.333 - 24 | = 5.667
(7.3)
Suppose that chocolate milk is favored by 48% of students over regular milk in a sample of 500 students. Use p̂ to estimate p. What is the probability that p̂ >.5? Make sure to check for normal
A) normal : np ≥ 10 n(1-p) ≥ 10
.48(500) = 240 ≥ 10 500(1-.48) = 260 ≥ 10
B) μp̂ = .48
σp̂ = √((.48(1-.48)/500) = .0223
normal cdf(area, 1E99, μp̂ , σp̂)
normalcdf( .5, 1E99, .48, .0223) = .1849
(7.3)
Anxiety affects 33% of teens. In a sample of 100 adolescents, let p denote the proportion of teens who have anxiety. find p̂, μp̂, and σp̂
p̂ = x/n = success over total
μp̂ = p̂
σp̂ = √((p(1-p)/n)
p̂ = 33/100 = .33
μp̂ = .33
σp̂ = √((.33(1-.33)/100) = .047
Other Quizlet sets
DHN 212 FINAL PREP
114 terms
GI, urinary, and reproductive systems
19 terms
Food Test
41 terms
HEED final
260 terms
Verified questions
STATISTICS
Which of the following situations involve permutations? I. 13 books are arranged on a shelf. II. 100 runners finish a race. III. An animal shelter chooses 3 kittens to put in the adopt-a-pet segment on the local news. A) I only B) II only C) I and II only D) I, II, and III
QUESTION
The following exercise refers to the small population {2, 6, 8, 10, 10, 12} with mean $$ \mu = 8 $$ and range 10. List all 15 possible SRSs of size n = 2 from the population. Find the value of $$ \overline{x} $$ for each sample.
STATISTICS
A probability experiment consists of rolling a 12-sided die. Find the probability of the event. Event C: rolling a number greater than 4
PROBABILITY
Consider the “hang time” punting data given in Case Study 12.2, using only the variables x2 and x3. (a) Verify the regression equation shown on page 489. (b) Predict punter hang time for a punter with LLS = 180 pounds and Power = 260 foot-pounds. (c) Construct a 95% confidence interval for the mean hang time of a punter with LLS = 180 pounds and Power = 260 foot-pounds.