Answer:
c) NORM.INV(R, 1200, 350)
Step-by-step explanation:
Given that the daily demand for gasoline at a local gas station is normally distributed with a mean of 1200 gallons, and a standard deviation of 350 gallons.
X = demand for gasolene at a local gas station is N(1200, 350)
R is any random number between 0 and 1.
Daily demand for gasolene would be
X = Mean + std deviation * z value, where Z = normal inverse of a value between 0 and 1.
The norm inv (R, 1200, 350) for R between 0 and 1 gives all the values of X
Hence correct choice would be
Option c) NORM.INV(R, 1200, 350)
Find the parametric equations for the line through the point P(2,4,4) that is perpendicular to the plane −1x+1y−4z=1. a) Use the variable t and write these equations so that t=0 corresponds to the point P.
Answer:
[tex]\frac{x-2}{-1} =\frac{y-4}{1} =\frac{z-4}{-4} =t[/tex]
Step-by-step explanation:
Given that a line passes through P(2,4,4)
Also the line is perpendicular to the plane
[tex]-1x+1y-4z=1.[/tex]
From the equation of the plane we can say that normal to the plane has direction ratios as (-1,1,-4)
Since the required line is also perpendicular to the plane, the direction ratios of the required line is
(-1,1,4)
It passes through (2,4,4)
If Q(x,y,z) are general points on the line then
Direction ratios of PQ are = (x-2, y-4, z-4)
These are proportional to (-1,1,4)
So parametric form of the line is
[tex]\frac{x-2}{-1} =\frac{y-4}{1} =\frac{z-4}{-4} =t[/tex]
Whem t=0 we get the point P.
Tri-Cities Bank has a single drive-in teller window. On Friday mornings, customers arrive at the drive-in window randomly, following a Poisson distribution at an average rate of 30 per hour.a. How many customers arrive per minute, on average?b. How many customers would you expect to arrive in a 10-minute interval?c. Use equation 13.1 to determine the probability of exactly 0, 1, 2, and 3 arrivals in a 10-minute interval. (You can verify your answers using the POISSON( ) function in Excel.)d. What is the probability of more than three arrivals occurring in a 10-minute interval?
Answer:
a) 0.5 per minutes
b) 5 arrivals expected in 10 minutes
c) P ( x = 0 ) = 0.00673 , P ( x = 1 ) = 0.03368 , P ( x = 2 ) = 0.08422 ,P ( x = 3 ) = 0.14037
d) P ( X >= 4 ) = 0.735
Step-by-step explanation:
Given:
- The number of customer arriving at window is modeled by Poisson distribution. The distribution is given by:
P(x) = ( λ^x ) (e^-λ) / x! x = 0 , 1 , 2 , 3 , ......
- Average rate λ = 30 / hr
Find:
a. How many customers arrive per minute, on average?
b. How many customers would you expect to arrive in a 10-minute interval?c. Use equation 13.1 to determine the probability of exactly 0, 1, 2, and 3 arrivals in a 10-minute interval.
d. What is the probability of more than three arrivals occurring in a 10-minute interval?
Solution:
- The average rate λ in number of customers that arrive in a minute is given by:
λ1 = 30 / 60 = 0.5 arrival per minutes
- The average number of customer that are expected to arrive in 10-minutes window is:
λ2 = 10*λ1 = 10*0.5 = 5 arrivals expected in 10 minutes
- The probability of exactly 0,1 , 2 , and 3 arrivals in 10 minute windows:
P ( x = 0 ) = ( 5^0 ) (e^-5) / 0! = 0.00673
P ( x = 1 ) = ( 5^1 ) (e^-5) / 1! = 0.03368
P ( x = 2 ) = ( 5^2 ) (e^-5) / 2! = 0.08422
P ( x = 3 ) = ( 5^3 ) (e^-5) / 3! = 0.14037
- The probability of more than three arrivals occuring in 10-minute interval is:
P ( X >= 4 ) = 1 - P ( X =< 3 )
P ( X >= 4 ) = 1 - [ P ( x = 0 ) + P ( x = 1 ) + P ( x = 2 ) + P ( x = 3 ) ]
P ( X >= 4 ) = 1 - [ 0.00673 + 0.03368 + 0.08422 + 0.14037 ]
P ( X >= 4 ) = 1 - [ 0.265 ]
P ( X >= 4 ) = 0.735
Using the Poisson distribution, it is found that:
a) 0.5 customers per minute.
b) 5 customers are expected to arrive.
c)
0.0068 = 0.68% probability of exactly 0 arrivals in a 10-minute interval.
0.0337 = 3.37% probability of exactly 1 arrivals in a 10-minute interval.
0.0842 = 8.42% probability of exactly 2 arrivals in a 10-minute interval.
0.1404 = 14.04% probability of exactly 3 arrivals in a 10-minute interval.
d) 0.7349 = 73.49% probability of more than three arrivals occurring in a 10-minute interval.
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by:
[tex]P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}[/tex]
The parameters are:
x is the number of successes e = 2.71828 is the Euler number [tex]\mu[/tex] is the mean in the given interval.Item a:
30 in one-hour(60 minutes), hence 0.5 customers per minute.
Item b:
0.5 customers per minute, hence, in a 10 minute interval, 5 customers are expected to arrive.
Item c:
[tex]P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-5}(5)^{0}}{(0)!} = 0.0068[/tex]
[tex]P(X = 1) = \frac{e^{-5}(5)^{1}}{(1)!} = 0.0337[/tex]
[tex]P(X = 2) = \frac{e^{-5}(5)^{2}}{(2)!} = 0.0842[/tex]
[tex]P(X = 3) = \frac{e^{-5}(5)^{3}}{(3)!} = 0.1404[/tex]
0.0068 = 0.68% probability of exactly 0 arrivals in a 10-minute interval.
0.0337 = 3.37% probability of exactly 1 arrivals in a 10-minute interval.
0.0842 = 8.42% probability of exactly 2 arrivals in a 10-minute interval.
0.1404 = 14.04% probability of exactly 3 arrivals in a 10-minute interval.
Item d:
This probability is:
[tex]P(X > 3) = 1 - P(X \leq 3)[/tex]
In which:
[tex]P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)[/tex]
From item c:
[tex]P(X \leq 3) = 0.0068 + 0.0337 + 0.0842 + 0.1404 = 0.2651[/tex]
Then:
[tex]P(X > 3) = 1 - P(X \leq 3) = 1 - 0.2651 = 0.7349[/tex]
0.7349 = 73.49% probability of more than three arrivals occurring in a 10-minute interval.
A similar problem is given at https://brainly.com/question/16912674
Consider the set consisting of the first 10 positive whole numbers (that is, 1-10). Complete parts (a) through (c) below.
(a) Determine explicitly the numbers in the set that are at least 4.
(b) Determine explicitly the numbers in the set that are at most 5
(c) Determine explicitly the numbers in the set that are between 4 and 6, inclusive.
Answer:
a) {4,5,6,7,8,9,10}
b) {1,2,3,4,5}
c) {4,5,6}
Research suggests that children who eat hot breakfast at home perform better at school. Many argue that not only hot breakfast but also parental care of children before they go to school has an impact on children's performance. In this case, parental care is: Group of answer choices An independent variable A dependent variable A mediating variable A moderating variable
Answer:children at home prefer hot breakfast than hot school breakfast because at home you could put it in the refrigerator but at school you have to throw it in the trash before you go to trash.
Step-by-step explanation:
Parental care in this context is considered the independent variable because it influences children's school performance.
Explanation:In this scenario, parental care is considered an independent variable. The reason for this is because it is the variable that influences or predicts the outcome, which in this case, is children's performance in school. The independent variable is the one that is manipulated or controlled in a study to observe its effects on the dependent variable (children's school performance here). Examples of parental care might include ensuring the child eats a good breakfast, aiding with schoolwork, or providing emotional support.
Learn more about Independent Variable here:https://brainly.com/question/32711473
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Seventy percent of children who go to the doctor have fevers. Of those with fevers, 30% also have a rash. Of those without fevers, 20% have a rash. What is the probability that a child at the doctor's office with a rash does not have a fever
Answer:
The probability that a child with a rash does not have a fever is 22%
Step-by-step explanation:
1. Probability of having fever:
[tex]P(fever)=0.70[/tex]
2. Probability of not having fever:
[tex]P(not fever)=1-P(fever)\\P(not fever)=1-0.70\\P(not fever)=0.30[/tex]
3. Probability of fave fevers and a rash:
[tex]P(fever and rash)=(0.70)(0.30)=0.21[/tex]
4. Probability of having a rash but not a fever:
[tex]P(rash and not fever)=(0.30)(0.20)=0.60[/tex]
5. Probability of having a rash:
[tex]P(rash)=P(rash and fever)+P(rash and no fever)\\P(rash)=0.21+0.06=0.27[/tex]
6. Probability a child with a rash does not have a fever
[tex]P=\frac{P(rash and not fever)}{P(rash)} =\frac{0.06}{0.27} =0.22[/tex]
22% of the child at the doctor's office with a rash does not have a fever.
Final answer:
To find the probability that a child with a rash does not have a fever, we analyze the given percentages, calculate how many children have a rash with and without fever, and then find the ratio of those without a fever to the total number with a rash, resulting in a probability of approximately 22.22%.
Explanation:
The question asks: What is the probability that a child at the doctor's office with a rash does not have a fever? To solve this, let's start by analyzing the given percentages.
70% of children who go to the doctor have fevers.
Of those with fevers, 30% also have a rash.
Of those without fevers, 20% have a rash.
To find the probability that a child with a rash does not have a fever, we need to calculate the proportion of children with a rash who are fever-free compared to all children with a rash.
Step-by-step Calculation:
Assuming 100 children visit the doctor: 70 will have fevers, and 30 will not.
Of the 70 with fevers, 21 (30% of 70) have a rash.
Of the 30 without fevers, 6 (20% of 30) have a rash.
In total, 27 children have a rash (21 with fever + 6 without).
The probability a child with a rash does not have a fever is the number of children with a rash but no fever divided by the total number of children with a rash: 6/27.
This calculation shows that the probability of a child having a rash but no fever is 6/27 or approximately 22.22%.
The Operations Manager likes your answer and thanks you for the analyses. He then asks you to do the ABC analysis. The manager adds that the company typically does ABC analysis such that the A segment (all A items togehter) accounts for at least 70% of value, B segment for at least 15% and C segment for at most 10% of value.
SKU Number of Items Value ($/item) Weight (lb/item)
A 220 67 52
B 225 3 59
C 245 21 60
D 145 7 48
E 230 26 42
F 240 83 49
Imagine, you want to first do the ABC analysis based on SKU value.
What percentage of the total number of items is accounted for by SKUs in segment B?
What percentage of the total value of inventory is accounted for by SKUs in segment B?
What percentage of the total weight of inventory is accounted for by SKUs in segment B?
Answer:
Requirements are computed below.
Step-by-step explanation:
For ABC analysis:
Step1: Calculate the total value of each item.
SKU Number of Items Value ($/item) Value of the product ($)
A 220 67 14740
B 225 3 675
C 245 21 5145
D 145 7 1015
E 230 26 5980
F 240 83 19920
Step2: Reorganize the table on the basis of total value in descending order, & calculate % share of total value of investment and cumulative % share of total value of investment.
SKU Number of Items Value ($/item) Value of the product ($) % of Total value Cumulative % of total value Segment
SKU Number of Items Value ($/item) Value of the product ($) % of Total value Cumulative % of total value Segment
F 240 83 19920 41.96 41.96 A
A 220 67 14740 31.05 73.01 A
E 230 26 5980 12.60 85.60 B
C 245 21 5145 10.84 96.44 B
D 145 7 1015 2.14 98.58 C
B 225 3 675 1.42 100.00 C
----------
% of total number of items in segment B = (230+245)/1305 = 36.4%
% of total value of inventory in segment B = (12.6 + 10.4)/100 = 23.43%
% of the total weight of inventory in segment B = (9660+14700)/67795 = 35.93%
SKU Number of Items Value ($/item) Weight (lb/item) Total weight
A 220 67 52 11440
B 225 3 59 13275
C 245 21 60 14700
D 145 7 48 6960
E 230 26 42 9660
F 240 83 49 11760
On July 1, 2015, Frank Corp. Purchased $100,000 of 8% bonds at face value. Interest is paid annually on June 30. If the accounting year for Frank ends at December 31, 2015, what will be reported with respect to the bonds on that date
Answer:
interest income in amount of $4000 will be accrued
Step-by-step explanation:
given data
principal = $100,000
rate = 8 percent = 0.08
time period = 6 month ( July - December ) = [tex]\frac{1}{2}[/tex] year
solution
we get here interest at December 31, 2015 that is express as
interest = principal × rate × time ..................1
put here value and we will get
interest = $100,000 × 0.08 × [tex]\frac{1}{2}[/tex]
interest = $100,000 × 0.08 × 0.5
interest = $4000
so interest income in amount of $4000 will be accrued
In Applied Life Data Analysis (Wiley, 1982), Wayne Nelson presents the breakdown time of an insulating fluid between electrodes at 34 kV. The times, in minutes, are as follows: 0.13, 0.68, 0.91, 1.36, 2.74, 3.08, 4.07, 4.71, 4.96, 6.56, 7.29, 7.91, 8.37, 12.11, 31.61, 32.65, 33.78, 36.72, and 72.96. Calculate the sample mean and sample standard deviation. Round the answers to 3 decimal places.
Answer:
Mean = 14.347 and Standard Deviation = 18.89
Step-by-step explanation:
Mean = Sum of all the numbers/total numbers = 272.6/19 = 14.347
For standard deviation, at first we will find variance which is
variance = The sum of the squared differences between each data point and the mean, divided by the number of data points (n) - 1:
x= data points
n= total number of data points
= ∑(x_i - Mean (x))^2/(n-1)
variance = 357
Standard deviation = [tex]\sqrt{variance}[/tex]
SD = [tex]\sqrt{357}[/tex]
SD = 18.89
Answer:
The mean= 14.347
The standard deviation =18.390
Step-by-step explanation:
The mean or average of a set of scores = the summation of the set of scores divided by the summation of the frequency of the respective scores.
Adding up the scores we have:
0.13+0.68+0.91+1.36+2.74+3.08+4.07+4.71+4.96+6.56+7.29+7.91+8.37+12.11+31.61+32.65+33.78+36.72+72.96 = 272.60
The frequency or number of scores is 19, so dividing this figure by the number of items in the set, we have:
272/19 =14.347
Therefore the mean of the scores (time, in minutes) is 14.347(to 3 decimal places)
In order to calculate the standard deviation of the set, we need to find the summation of the squares of the different deviations from the mean.
Example, 0.13 is the first score in the data. We subtract the mean (14.35) from 0.13 and we'll have 14.22 which is the score's deviation from the mean. the next step is to find the square of the deviation,14.22 which will be 202.208.
We'll repeat this same process for the remainder of the scores and then sum up the squares of the deviations.Doing this, the summation of the squares of all the deviations from the mean score will be = 6425.67
Once this is calculated we then solve to obtain the standard deviation of the scores by applying the formula:
√summation of (x - mean deviations)/total number of scores
=√(6425.67/19)
= √ (338.193)
= 18.390( to 3 decimal places)
Therefore the mean and the standard deviation of the set of scores are 14.347 and 18.390 respectively
During the period of time that a local university takes phone-in registrations, calls come in at the rate of one every two minutes.
What is the expected number of calls in one hour?
Answer:
Let X the random variable who represent the number of occurences in a period of time for the calls.
For this case we have the following parameter [tex] \lambda = 1 \frac{call}{2 minutes}[/tex]
And we are interested in the expected number of calls in one hour.
We know that 1 hr = 60 mins so then the expected number of calls that arrive in one hour are:
[tex] \lambda = \frac{1 call}{2 minutes} * \frac{60 minutes}{1 hour} = 30 calls per hour[/tex]
Step-by-step explanation:
Definitions and concepts
The Poisson process is useful when we want to analyze the probability of ocurrence of an event in a time specified. The probability distribution for a random variable X following the Poisson distribution is given by:
[tex]P(X=x) =\lambda^x \frac{e^{-\lambda}}{x!}[/tex]
And the parameter [tex]\lambda[/tex] represent the average ocurrence rate per unit of time.
Solution to the problem
Let X the random variable who represent the number of occurences in a period of time for the calls.
For this case we have the following parameter [tex] \lambda = 1 \frac{call}{2 minutes}[/tex]
And we are interested in the expected number of calls in one hour.
We know that 1 hr = 60 mins so then the expected number of calls that arrive in one hour are:
[tex] \lambda = \frac{1 call}{2 minutes} * \frac{60 minutes}{1 hour} = 30 calls per hour[/tex]
A can in the shape of a right circular cylinder is required to have a volume of 700 cubic centimeters. The top and bottom are made up of a material that costs 8� per square centimeter, while the sides are made of material that costs 5� per square centimeter. Find a function that describes the total cost of the material as a function of the radius r of the cylinder
Answer:
the cost function is Cost=7000 m*$ /R + 50.265 $/m² * R²
Step-by-step explanation:
then the cost function is
Cost= cost of side area+ cost of top + cost of bottom = 2*π*R*L * 5$/m² +
π*R² * 8$/m² + π*R² * 8$/m²
since the volume V is
V=π*R²*L → V/(π*R²)=L
then
Cost=2*π*R*V/(π*R²) * 5$/m² + π*R² * 8$/m² + π*R² * 8$/m²
replacing values
Cost=2*700 m³ /R * 5$/m² + π*R² * 16$/m² = 7000 m*$ /R + 50.265 $/m² * R²
thus the cost function is
Cost=7000 m*$ /R + 50.265 $/m² * R²
Answer: 50.24r² + 7000/r
Step-by-step explanation:
The formula for determining the volume of a cylinder is expressed as
Volume = πr²h
Since the volume of the can is 700cm³, then
πr²h = 700
h = 700/πr²
The formula for determining the total surface area of a cylinder is expressed as
Total surface area = 2πr² + 2πrh
The surface area of the top and bottom of the can is 2πr².
Since top and bottom are made up of a material that costs $8 per square centimeter, then the cost is
2πr² × 8 = 16πr²
Since π = 3.14, the surface area of the top and bottom of the cylindrical can is
16 × 3.14 × r² = 50.24r²
The surface area of the side of the can is
2πrh = 2πr × 700/πr²
= 1400/r
Since the the sides are made of material that costs $5 per square centimeter, then the cost is
1400/r × 5 = 7000/r
The total cost of the material as a function of the radius, r of the cylinder is
50.24r² + 7000/r
How much time do Americans spend eating or drinking? Suppose for a random sample of 1001 Americans age 15 or older, the mean amount of time spent eating or drinking per day is 1.22 hours with a standard deviation of 0.65 hour. (a) A histogram of time spent ea ting and drinking each day is skewed right. Use this result to explain why a large sample size is needed to construct a confidence interval for the mean time spent eating and drinking each day. (b) There are over 200 million Americans age 15 or older. Explain why this, along with the fact that the data were obtained using a random sample, satisfies the requirements for constructing a confidence interval. (c) Determine and interpret a 95% confidence interval for the mean amount of time Americans age 15 or older spend eating and drinking each day. (d) Could the interval be used to estimate the mean amount of time a 9-year-old American spends eating and drinking each day? Explain.
Answer:
Step-by-step explanation:
a)
To construct a confidence interval ,the histogram should be bell shaped.It means that the it should be normally distributed with no skewness.To eliminate the skewness a large sample size is required so that the sample is normally distribute about the mean Also ,in order to construct a t-interval the sample data must come from the population that is normally distributed or the sample size is larger than 30.Since the question stated that the population distribution is skewed to the right \bar{x} is guaranteed to be normally distributed if [tex]n\geq 30[/tex]
b)The sample satisfies the normal distribution because the sample sixe is greater than 30 which is 1001.
c) [[tex]\bar{x}=1.22 \,s=.65\, n=1001[/tex]
Area in the right tail =2.5% or .025
degree of freedom =1001-1 =1000
[tex]t_{\alpha /2}=1.96[/tex]
95% confidence interval is :[tex]\bar{x}\pm t_{\alpha /2}s/\sqrt{n} [/tex]
lower bound is :[tex]1.22-1.96*.65/\sqrt{1001} =1.18 [/tex]
upper bound is :[tex]-1.22+1.96*.65/\sqrt{1001}=1.26 [/tex]
d)No because the sample data was obtined from Americans age 15 or older which cannot be applied to a different age group
The mayor of a town has proposed a plan for the annexation of a new community. A political study took a sample of 10001000 voters in the town and found that 56V% of the residents favored annexation. Using the data, a political strategist wants to test the claim that the percentage of residents who favor annexation is more than 53S%. Find the value of the test statistic. Round your answer to two decimal places.
Answer:
Test statistics = 1.87
Step-by-step explanation:
We are given that a political study took a sample of 1000 voters in the town and found that 56% of the residents favored annexation.
And, a political strategist wants to test the claim that the percentage of residents who favor annexation is more than 53%, i.e;
Null Hypothesis, [tex]H_0[/tex] : p = 0.53 {means that the percentage of residents who favor annexation is 53%}
Alternate Hypothesis, [tex]H_1[/tex] : p > 0.53 {means that the percentage of residents who favor annexation is more than 53%}
The test statistics we will use here is;
T.S. = [tex]\frac{\hat p -p}{\sqrt{\frac{\hat p(1- \hat p)}{n} } }[/tex] ~ N(0,1)
where, p = actual percentage of residents who favor annexation = 0.53
[tex]\hat p[/tex] = percentage of residents who favor annexation in a sample of
1000 voters = 0.56
n = sample of voters = 1000
So, Test statistics = [tex]\frac{0.56 -0.53}{\sqrt{\frac{0.56(1- 0.56)}{1000} } }[/tex]
= 1.87
Therefore, the value of test statistics is 1.87 .
The alkalinity level of water specimens collected from the Han River in Seoul, Korea, has a mean of 50 milligrams per liter and a standard deviation of 3.2 milligrams per liter. (Environmental Science & Engineering, Sept. 1, 2000.) Assume the distribution of alkalinity levels is approximately normal and find the probability that a water specimen collected from the river has an alkalinity level a. exceeding 45 milligrams per liter. b. below 55 milligrams per liter. c. between 48 and 52 milligrams per liter.
Answer:
a) 94.06% probability that a water specimen collected from the river has an alkalinity level exceeding 45 milligrams per liter.
b) 94.06% probability that a water specimen collected from the river has an alkalinity level below 55 milligrams per liter.
c) 50.98% probability that a water specimen collected from the river has an alkalinity level between 48 and 52 milligrams per liter.
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
[tex]\mu = 50, \sigma = 3.2[/tex]
a. exceeding 45 milligrams per liter.
This probability is 1 subtracted by the pvalue of Z when X = 45. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{45 - 50}{3.2}[/tex]
[tex]Z = -1.56[/tex]
[tex]Z = -1.56[/tex] has a pvalue of 0.0594.
1 - 0.0594 = 0.9406
94.06% probability that a water specimen collected from the river has an alkalinity level exceeding 45 milligrams per liter.
b. below 55 milligrams per liter.
This probability is the pvalue of Z when X = 55.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{55 - 50}{3.2}[/tex]
[tex]Z = 1.56[/tex]
[tex]Z = 1.56[/tex] has a pvalue of 0.9604.
94.06% probability that a water specimen collected from the river has an alkalinity level below 55 milligrams per liter.
c. between 48 and 52 milligrams per liter.
This is the pvalue of Z when X = 52 subtracted by the pvalue of Z when X = 48. So
X = 52
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{52 - 50}{3.2}[/tex]
[tex]Z = 0.69[/tex]
[tex]Z = 0.69[/tex] has a pvalue of 0.7549
X = 48
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{48 - 50}{3.2}[/tex]
[tex]Z = -0.69[/tex]
[tex]Z = -0.69[/tex] has a pvalue of 0.2451
0.7549 - 0.2451 = 0.5098
50.98% probability that a water specimen collected from the river has an alkalinity level between 48 and 52 milligrams per liter.
The probabilities are calculated using the normal distribution properties: exceeding 45 mg/L is approximately 0.9406, below 55 mg/L is approximately 0.9406, and between 48 and 52 mg/L is approximately 0.4678.
a) To find the probability that the alkalinity level exceeds 45 milligrams per liter, we can use the standard normal distribution with the given mean (50 mg/L) and standard deviation (3.2 mg/L). First, we need to calculate the z-score for 45 mg/L:
z = 45 - 50/3.2 = -1.5625
Then, we find the probability of the alkalinity level exceeding 45 mg/L by finding the area to the right of this z-score in the standard normal distribution. Using a standard normal table or calculator, we find this probability to be approximately 0.9406.
b) Similarly, to find the probability that the alkalinity level is below 55 milligrams per liter, we calculate the z-score for 55 mg/L:
[tex]\[ z = \frac{55 - 50}{3.2} = 1.5625 \][/tex]
Then, we find the probability of the alkalinity level being below 55 mg/L by finding the area to the left of this z-score in the standard normal distribution. Using a standard normal table or calculator, we find this probability to be approximately 0.9406.
c) To find the probability that the alkalinity level is between 48 and 52 milligrams per liter, we first calculate the z-scores for these values:
z₁ = 48-50/3.2 = -0.625
z₂ = 52-50/3.2 = 0.625
Then, we find the area between these two z-scores in the standard normal distribution, which represents the probability of the alkalinity level being between 48 and 52 mg/L. Using a standard normal table or calculator, we find this probability to be approximately 0.3146.
In conclusion:
Probability exceeding 45 mg/L: ≈ 0.9406Probability below 55 mg/L: ≈ 0.9406Probability between 48 and 52 mg/L: ≈ 0.4678.A coffee company wants to make sure that their coffee is being served at the right temperature. If it is too hot, the customers could burn themselves. If it is too cold, the customers will be unsatisfied. The company has determined that they want the average coffee temperature to be 65 degrees C. They take a sample of 20 orders of coffee and find the sample mean to be equal to 70.2 C. What does mu represent for this problem?
In the given sample, the parameter [tex]\mu[/tex] represents the average temperature of the coffee in the population whose value is not known.
In statistics, a parameter means a numerical attribute of a population. A population parameter describes a particular aspect of the entire population, which is the complete set of individuals, items, or units of interest. The mean, standard deviation, and variance are some of the population parameters.
A coffee company wants to make sure that their coffee is being served at the right temperature. If it is too hot, the customers could burn themselves. If it is too cold, the customers will be unsatisfied. The company has determined that they want the average coffee temperature to be 65 degrees C. They take a sample of 20 orders of coffee and find the sample mean to be equal to 70.2 C. [tex]\mu[/tex] represents the average temperature of the coffee in the population, and the value is not known for this problem.
Also, [tex]\bar{X}[/tex] represents the sample mean here which is known and equals 70.2 C.
b
Therefore, [tex]\mu[/tex] represents the population mean and is not known in this problem.
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The symbol mu in the coffee company's problem represents the population mean, the desired average coffee temperature, which should be 65 degrees C.
In the context of the coffee company's quality control problem, the symbol mu represents the population mean, which is the average temperature of all cups of coffee served by the company. The sample mean, denoted as x-bar, is 70.2°C, which is the average temperature calculated from the sample of 20 orders of coffee. The goal of the company is to have mu equal to 65°C, as they want this to be the average serving temperature to ensure customer satisfaction and safety. If they find that x-bar is significantly different from mu, it may suggest that corrective actions are needed to reach the desired temperature.
Aimee sells hand-embroidered dog apparel over the Internet. Her annual revenue is $128,000 per year, the explicit costs of her business are $42,000, and the opportunity costs of her business are $30,000. What are the implicit costs of her business?
Answer:
200,000
Step-by-step explanation:
add 128,000+42,000+30,000
The implicit cost of her business is $56000
What is the implicit cost?Implicit costs are a specific type of opportunity cost, the cost of resources already owned by the firm that could have been put to some other use.
Given that, the annual revenue of Aimee is $128,000, the explicit costs of her business are $42,000, and the opportunity costs of her business are $30,000.
We need to find the implicit costs of her business,
Therefore,
Economic profit = total revenues – explicit costs – implicit costs.
30000 = 128000-42000-implicit costs.
Implicit costs = 56000
Hence, the implicit cost of her business is $56000
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Elise’s garden is below. Tomato plants require 2 square feet of space.
(Shape of garden is a Rectangle 6.3 feet by 12.6 feet )
How many tomato plants can Elise fit in her garden?
Answer:
Step-by-step explanation:
You should try going too jeeska
Answer:
Step-by-step explanation:
Elise’s garden is rectangular. The formula for determining the area of a rectangle is expressed as
Area = length × width
From the information given,
Length of garden = 6 feet
Width of garden = 12.6 feet
Area of garden = 6 × 12.6 = 75.6 square feet
Tomato plants require 2 square feet of space. This means that the number of tomato plants that Elise can fit in her garden is
75.6/2 = 37.8
Since the number of tomato plants must be whole number, then the number of tomato plants that Elise can fit in her garden is 37
Testing for a Vector Space In Exercises 13–36, determine whether the set, together with the standard operations, is a vector space. If it is not, identify at least one of the ten vector space axioms that fails. 13. M, 4.6 14. M, 15. The a set of all third-degree polynomials 16. The set of all fifth-degree polynomials 17. The set of all first-degree polynomial functions ax, a t 0, whose graphs pass through the origin 18. The set of all first-degree polynomial functions ax + b a, b 0, whose graphs do not pass through the origin 19. The set of all polynomials of degree four or less
Answer:
Step-by-step explanation:
13. The set M4,6 of all 4x6 matrices is a vector space as it is closed under vector addition as also scalar multiplication. Also the 4x6 zero matrix is in M4,6.
14. The set M1,1 is a singleton, i.e. a 1x1 matrix. It is a vector space for the same reasons as in 13. above.
15. The degree of the zero polynomial is undefined and it is usually treated as a constant (of degree 0). If we treat 0 as a polynomial of degree 0, then the set of all 3rd degree polynomials is not a vector space despite being closed under vector addition and scalar multiplication as it does not contain the zero polynomial.
16. The set of all 5th degree polynomials is not a vector space despite being closed under vector addition and scalar multiplication as it does not contain the zero polynomial.
17. The set of all first degree polynomials ax (a≠0) is not a vector space. If p(x) = ax and q(x) = -ax , then p(x)+q(x) = 0*x which is not in the given set. Hence the set is not closed under vector addition and, therefore, it is not a vector space.
18. The set of all first degree polynomials ax +b (a,b ≠0) is not a vector space. If p(x) = ax +b and q(x) = -ax -b , then p(x)+q(x) = 0*x +0 which is not in the given set. Hence the set is not closed under vector addition and, therefore, it is not a vector space.
19. The set P4 of all polynomials of degree 4 or less isa vector space. If p(x) = a1x4+a2 x3+a3x2+a4x+a5 and q(x) = b1x4+b2 x3+b3x2+b4x+b5 are 2 arbitrary elements of P4, then p(x)+q(x)is in P4. Similarly, αp(x) is in P4 for any arbitrat scalar α. Hence, P4 is closed under vector addition and scalar multiplication. Also, the 0 polynomial is in P4. Hence P4 is a vector space.
A set forms a vector space if it meets all ten vector space axioms. For instance, a set of first-degree polynomial functions ax + b, where a, and b are non-zero and whose graphs do not pass through the origin, fails to meet vector space axiom for closure under addition. On the other hand, the set of all third-degree polynomials is a vector space as it satisfies all ten vector space axioms.
Explanation:To determine whether a set combined with standard operations forms a vector space, it must satisfy ten vector space axioms. These axioms include requirements regarding the addition and scalar multiplication of vectors.
To illustrate, let's look at the set of all first-degree polynomial functions ax + b, with a, b ≠ 0, whose graphs do not pass through the origin. By definition, vectors in a vector space should be closed under addition, i.e., if you add any two vectors together, you should get another vector in the set. However, if we add two such functions together ax + b + cx + d = (a+c)x + (b+d), the resulting function will pass through the origin only if (b+d)=0. But given that b and d are non-zero, the sum will not yield a first-degree polynomial that passes through the origin. Therefore, this set fails to meet the vector space axiom for closure under addition.
On the other hand, the set of all third-degree polynomials will be a vector space as it adheres to all the ten axioms. For example, if two polynomials in the set are added together or multiplied by a scalar, the resultant is still a third-degree polynomial which means it still belongs to the set.
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New York City is one of the most expensive cities in the US for lodging. The mean hotel room rate is $244.00 per night; assume that room rates are normally distributed, with the standard deviation of $55.00 What is the probability that a hotel room costs between $250.00 and $285.00?
Answer:
22.96% probability that a hotel room costs between $250.00 and $285.00
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
[tex]\mu = 244, \sigma = 55[/tex]
What is the probability that a hotel room costs between $250.00 and $285.00?
This is the pvalue of Z when X = 285 subtracted by the pvalue of Z when X = 250. So
X = 285
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{285 - 244}{55}[/tex]
[tex]Z = 0.75[/tex]
[tex]Z = 0.75[/tex] has a pvalue of 0.7734
X = 250
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{250 - 244}{55}[/tex]
[tex]Z = 0.11[/tex]
[tex]Z = 0.11[/tex] has a pvalue of 0.5438
0.7734 - 0.5438 = 0.2296
22.96% probability that a hotel room costs between $250.00 and $285.00
A professor planned to give an examination in a large class on the Monday before Thanksgiving vacation. Some students asked whether he could change the date because so many of their classmates had at least one other exam on that date. They speculated that at least 40% of the class had this problem. The professor agreed to poll the class, and if there was convincing evidence that the proportion with at least one other exam on that date was greater than .40, he would change the date. Of the 250 students in the class, 109 reported that they had another exam on that date. What proportion of the class reported that they had another exam on the date?
Answer:
0.436
Step-by-step explanation:
Given that a professor planned to give an examination in a large class on the Monday before Thanksgiving vacation
Some students asked whether he could change the date because so many of their classmates had at least one other exam on that date.
They speculated that at least 40% of the class had this problem.
The professor agreed to poll the class, and if there was convincing evidence that the proportion with at least one other exam on that date was greater than .40, he would change the date.
No of students in total = 250
Reported they had another exam = 109
proportion of the class reported that they had another exam on the date
= [tex]\frac{109}{250} =0.436[/tex]
Sandra just finished planting avacados,
carrots, radishes, tomatoes, and spinach in her new garden. The
garden is a circle whose diameter is 50 yards. If she planted equal
regions of each vegetable, what is the area of Sandra's garden that
has carrots? .
Answer:392.5
Step-by-step explanation:
50/2=25
25*25=625
625*3.14=1962.5
1962.5/5=392.5
Im right look at pic for proof
Translate the statement into a confidence interval for p. Approximate the level of confidence.
In a survey of 1000 U.S. adults, 19% are concerned that their taxes will be audited by the Internal Revenue Service. The survey's margin of error is plus or minus 3%. (Source: Rasmussen Reports)
Approximate the level of confidence.
The interval is approximately a _______% confidence interval.
Answer:
The interval is approximately a 98% confidence interval
Step-by-step explanation:
From the question : Error, E= 3% = 0.03, Total population, n=1000, number of people that are concerned that their taxes will be audited, p = 19% = 0.19
E^{2}=z_{\alpha/2}^{2}\cdot \frac{p(1-p)}{n}
0.03^{2}=z_{\alpha/2}^{2}\cdot \frac{0.19(1-0.19)}{1000}
z_{\alpha/2}^{2}=5.848
z_{\alpha/2}=2.418
Area right to 2.418 is 0.0078. So
\alpha/2=0.0078
Therefore \alpha=0.0156\approx 0.02
Thus, the interval is approximately a 98% confidence interval.
Answer:
98 percent
Step-by-step explanation:
plato
Read the paragraph and answer the question below: Results from two CNN/USA Today/Gallup polls, one conducted in March 2003 and one in November 2003, were recently presented online. Both polls involved samples of 1001 adults, aged 18 years and older. In the March sample, 45% of those sampled claimed to be fans of professional baseball whereas 51% of those polled in November claimed to be fans. Construct a 99% confidence interval for the proportion of adults who professed to be baseball fans in November 2003 (after the World Series). Interpret this interval.
Answer:
99% confidence interval for the proportion of adults who professed to be baseball fans in November 2003 is (46.9%, 55.1%)
This interval means that the lower limit of the proportion of adults who professed to be baseball fans is 46.9% and the upper limit of the proportion is 55.1%
Step-by-step explanation:
Confidence interval = P' + or - t×sqrt[P'(1-P') ÷ n]
P' is sample proportion = 51% = 0.51
n = 1001
confidence level = 99%
t-value corresponding to 99% confidence interval and infinity degree of freedom is 2.576
t × sqrt[P'(1-P') ÷ n] = 2.576 × sqrt[0.51(1-0.51) ÷ 1001] = 2.576 × 0.0158 = 0.041
Lower limit = P' - 0.041 = 0.51 - 0.041 = 0.469 = 46.9%
Upper limit = P' + 0.041 = 0.51 + 0.041 = 0.551 = 55.1%
99% Confidence interval is (46.9%, 55.1%)
The interval means that the proportion of adults who professed to be baseball fans in November 2003 is between 46.9% and 55.1%
You have $3,500 on a credit card that charges a 13% interest rate. If you want to pay off the credit card in 4 years, how much will you need to pay each month (assuming you don't charge anything new to the card)?
To pay off your credit card debt of $3,500 at 13% interest rate in 4 years, you will have to calculate the monthly payments by using a certain formula. After defining the parameters (monthly payment, monthly interest rate, present value or current loan amount, and time), all you have to do is put the values in it and calculate the monthly payment.
Explanation:This question refers to a financial problem in the subject of Mathematics, particularly pertaining to the concept of
simple interest
. To find out how much you need to pay each month to clear out your credit card debt, you would need to understand how the 13% interest is applied to your debt of $3,500. The formula for finding monthly payments is
P = [r*PV(1 + r)^t]/[(1 + r)^t – 1]
, where:
P is your monthly payment,r is your monthly interest rate, calculated by taking your annual rate (0.13 in this case) and dividing it by 12,PV is your present value or current loan amount, which is $3500 in this case, andt is time, the number of periods in which payments will be made, which is 4*12 = 48 months in this case.After defining the parameters, all you have to do is put the values in the formula and calculate the monthly payment.
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To calculate the monthly payments on a credit card debt of $3,500 with a 13% interest rate over 4 years, you can use the formula for calculating the monthly payment on a fixed-rate loan. The monthly payment comes out to be approximately $94.78.
Explanation:To calculate the monthly payments on a credit card debt, you can use the formula for calculating the monthly payment on a fixed-rate loan. The formula is:
Monthly Payment = (Principal * Monthly Interest Rate) / (1 - (1 + Monthly Interest Rate)^(-Number of Months))
In this case, the principal is $3,500, the monthly interest rate is 13% divided by 12 (0.13/12), and the number of months is 4 years multiplied by 12 months (4 * 12). Plugging in these values, we get:
Monthly Payment = (3500 * 0.13/12) / (1 - (1 + 0.13/12)^(-4 * 12))
Solving this equation will give us the monthly payment amount. Using a calculator, the monthly payment comes out to be approximately $94.78.
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It is known that 70% of all brand A external hard drives work in a satisfactory manner throughout the warranty period (are "successes"). Suppose that n = 15 drives are randomly selected. Let X = the number of successes in the sample. The statistic X/n is the sample proportion (fraction) of successes. Obtain the sampling distribution of this statistic. [Hint: One possible value X/n is 0.2, corresponding to X = 3. What is the probability of this value (what kind of random variable is X)?] (Round your answers to three decimal places.)
Answer:
The probability distribution of [tex]\hat p[/tex] is, [tex]N(0.70, 0.032)[/tex].
Step-by-step explanation:
The random variable X is defined as the number of brand A external hard drives work in a satisfactory manner throughout the warranty period.
The sample selected is of size, n = 15.
The probability of selecting a hard drive that works in a satisfactory manner throughout is, p = 0.70.
Every hard drive works independently of the others.
The random variable X follows a Binomial distribution with parameters n = 15 and p = 0.70.
The probability mass function of X, the binomial random variable is:
[tex]P(X=x)={15\choose x}0.70^{x}(1-0.70)^{15-x};\ x=0,1,2,3...[/tex]
The probability distribution of X is shown below.
Now a statistic is defined as:
[tex]\hat p=\frac{X}{n}[/tex]
This statistic is known as the sample proportion.
A Normal approximation to Binomial can be used to approximate the distribution of sample proportion if the following conditions are satisfied:
np ≥ 5n(1 - p) ≥ 5Then the sampling distribution of [tex]\hat p[/tex] follows a Normal; distribution with mean [tex]p=0.70[/tex] and standard deviation [tex]\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.70(1-0.70)}{200}}=0.032[/tex].
Check the conditions:
[tex]np=15\times0.70=10.5\\n(1-p)=15\times (1-0.70)=4.5\approx5[/tex]
So, the probability distribution of [tex]\hat p[/tex] is, [tex]N(0.70, 0.032)[/tex].
The sampling distribution of X/n is approximately normally distributed with a mean of 0.7 and a standard deviation of 0.117.
To determine the sampling distribution of the statistic X/n, we need to consider the probability distribution of the number of successes (X) in the sample of n = 15 drives.
Since the probability of a success (a working hard drive) is p = 0.7 and the probability of a failure (a non-working hard drive) is q = 1 - p = 0.3, we can model X using the binomial distribution with parameters n = 15 and p = 0.7.
The probability of getting exactly x successes (working hard drives) in a sample of n = 15 drives is given by the binomial probability mass function (PMF):
P(X = x) = (nCx) * px * (1-p)^(n-x)
where:
nCx is the binomial coefficient, which represents the number of ways to choose x successes out of n trials.
px is the probability of success (working hard drive) on each trial.
(1-p)^(n-x) is the probability of failure (non-working hard drive) on each trial.
For x = 3 successes (working hard drives), we calculate the probability:
P(X = 3) = (15C3) * (0.7)^3 * (0.3)^12 ≈ 0.193
This means that the probability of observing 3 successes (working hard drives) in a sample of 15 drives is approximately 0.193.
The sample proportion (fraction) of successes, X/n, is a continuous random variable that can take values between 0 and 1. To determine the sampling distribution of X/n, we need to consider the distribution of X and how it transforms to X/n.
The distribution of X/n is approximately normal for large sample sizes (n ≥ 30), but for smaller sample sizes, it can be skewed. In this case, with n = 15, the distribution of X/n is slightly skewed, but it is still close to being normal.
The mean of the sampling distribution of X/n is equal to the population proportion, which is p = 0.7.
The standard deviation of the sampling distribution of X/n is given by:
σ(X/n) = √(p * q / n) ≈ √(0.7 * 0.3 / 15) ≈ 0.117
Therefore, the sampling distribution of X/n is approximately normally distributed with a mean of 0.7 and a standard deviation of 0.117.
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Please help!!!!!! I’ll mark you as brainliest if correct
Answer: 771,243
Step-by-step explanation:
Is (2,7) a point on the line y=4x-3?
YES OR NO?
Answer:
mmmm
yesssirrreeee
A building engineer analyzes a concrete column with a circular cross section. The circumference of the column is 18 \pi18π18, pi meters.
What is the area AAA of the cross section of the column?
Give your answer in terms of pi.
The area of the cross section of the column is [tex]18 \pi \ m^2[/tex]
Explanation:
Given that a building engineer analyzes a concrete column with a circular cross section.
Also, given that the circumference of the column is [tex]18 \pi[/tex] meters.
We need to determine the area of the cross section of the column.
The area of the cross section of the column can be determined using the formula,
[tex]Area= \pi r^2[/tex]
First, we shall determine the value of the radius r.
Since, given that circumference is [tex]18 \pi[/tex] meters.
We have,
[tex]2 \pi r=18 \pi[/tex]
[tex]r=9[/tex]
Thus, the radius is [tex]r=9[/tex]
Now, substituting the value [tex]r=9[/tex] in the formula [tex]Area= \pi r^2[/tex], we get,
[tex]Area = \pi (9)^2[/tex]
[tex]Area = 81 \pi[/tex]
Thus, the area of the cross section of the column is [tex]18 \pi \ m^2[/tex]
Trapezoid ABCD is rotated on 180° about the origin. Draw the image A'B'C'D' of the given trapezoid and determine which of the following statements are true?
Select all correct statements.
A. A'D' ∥B'C'
B. A'B' ∥ D'C'
C. AD∥ A'D'
D. A'B' ∥ AB
E. AD ∥ B'C'
Answer: B, C, D
Step-by-step explanation:
since it is rotated, the parallel sides stay parallel.
The trapezoid ABCD rotated to form A'B'C'D, A'B' ∥ D'C',
TransformationTransformation is the movement of a point from its initial location to a new location. Types of transformation are rotation, reflection, dilation and translation.
Rotation is a rigid transformation, hence it preserves the shape and size. If a point A(x, y) is rotated on 180° about the origin, the new point is A'(-x, -y).
Hence for the trapezoid ABCD rotated to form A'B'C'D, A'B' ∥ D'C',
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A manufacturer of nickel-hydrogen batteries randomly selects 100 nickel plates for test cells, cycles them a specified number of times and determines that 14 of the plates have blistered.
Does this data provide compelling evidence for concluding that more than 10% of all plates blister under such circumstances?
Use Alpha =0.10.
A. What is the parameter of interest?
B. State the null and alternative hypotheses.
C. Calculate the test statistic.
D. Find the rejection region.
E. Make a decision and interpret.
F. Find a p-value corresponding to the test and compare with your decision in E.
Answer:
a) Parameter of interest [tex] p[/tex] representing the true proportion of the plates have blistered.
b) Null hypothesis:[tex]p\leq 0.1[/tex]
Alternative hypothesis:[tex]p > 0.1[/tex]
c) [tex]z=\frac{0.14 -0.1}{\sqrt{\frac{0.1(1-0.1)}{100}}}=1.33[/tex]
d) For this case we need to find a value in the normal standard distribution that accumulates 0.1 of the area in the right tail and for this case is:
[tex] z_{critc}= 1.28[/tex]
e) For this case since our calculated value is higher than the critical value 1.33>1.28 we have enough evidence to reject the null hypothesis and we can conclude that the true proportion is significantly higher than 0.1
f) [tex]p_v =P(z>1.33)=0.0917[/tex]
If we compare the p value and the significance level given [tex]\alpha=0.1[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 10% of significance the true proportion is higher than 0.1 or 10%
Step-by-step explanation:
Data given and notation
n=100 represent the random sample taken
Part a
Parameter of interest [tex] p[/tex] representing the true proportion of the plates have blistered.
X=14 represent the number of the plates have blistered.
[tex]\hat p=\frac{14}{100}=0.14[/tex] estimated proportion of the plates have blistered.
[tex]p_o=0.1[/tex] is the value that we want to test
[tex]\alpha=0.1[/tex] represent the significance level
Confidence=90% or 0.90
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value (variable of interest)
Part b: Concepts and formulas to use
We need to conduct a hypothesis in order to test the claim that more than 10% of all plates blister under such circumstances.:
Null hypothesis:[tex]p\leq 0.1[/tex]
Alternative hypothesis:[tex]p > 0.1[/tex]
When we conduct a proportion test we need to use the z statistic, and the is given by:
[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)
The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].
Part c: Calculate the statistic
Since we have all the info requires we can replace in formula (1) like this:
[tex]z=\frac{0.14 -0.1}{\sqrt{\frac{0.1(1-0.1)}{100}}}=1.33[/tex]
Part d: Rejection region
For this case we need to find a value in the normal standard distribution that accumulates 0.1 of the area in the right tail and for this case is:
[tex] z_{critc}= 1.28[/tex]
Part e
For this case since our calculated value is higher than the critical value 1.33>1.28 we have enough evidence to reject the null hypothesis and we can conclude that the true proportion is significantly higher than 0.1
Part f
Since is a right taild test the p value would be:
[tex]p_v =P(z>1.33)=0.0917[/tex]
If we compare the p value and the significance level given [tex]\alpha=0.1[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 10% of significance the true proportion is higher than 0.1 or 10%
The brain volumes (cm cubed) of 50 brains vary from a low of 902 cm cubed to a high of 1494 cm cubed. Use the range rule of thumb to estimate the standard deviation s and compare the result to the exact standard deviation of 183.2 cm cubed, assuming the estimate is accurate if it is within 15 cm cubed.
Answer:
A) Estimated standard deviation = 148cm³
B) The estimated standard deviation of 148 cm³ is less than the true standard deviation of 183.2 cm³. This estimated standard deviation is not even within 15 cm³ of the true standard deviation and thus we can say it's not accurate.
Step-by-step explanation:
From the question, the given standard deviation is 183.2 cm³
Also that the range of values is between 902 cm³ to 1494 cm³.
Range is the difference between highest and lowest values in the data set.
Thus, Range = 1494 cm³ - 902 cm³ = 592 cm³
Now, The range rule tells us that the standard deviation of a sample is approximately equal to one-fourth of the range of the data. In other words s = (Maximum – Minimum)/4
Thus, estimated standard deviation = 592/4 = 148 cm³