Answer:
The 80% confidence interval for the average net change is (8.596, 12.904).
Critical value t=1.638.
Step-by-step explanation:
First, we calculate the mean and standard deviation of the sample:
[tex]M=\dfrac{1}{n}\sum_{i=1}^n\,x_i\\\\\\M=\dfrac{1}{4}(12+7+13+11)\\\\\\M=\dfrac{43}{4}\\\\\\M=10.75\\\\\\s=\dfrac{1}{n-1}\sum_{i=1}^n\,(x_i-M)^2\\\\\\s=\dfrac{1}{3}((12-10.75)^2+(7-10.75)^2+(13-10.75)^2+(11-10.75)^2)\\\\\\s=\dfrac{20.75}{3}\\\\\\s=6.92\\\\\\[/tex]
We have to calculate a 80% confidence interval for the mean.
The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.
The sample mean is M=10.75.
The sample size is N=4.
When σ is not known, s divided by the square root of N is used as an estimate of σM:
[tex]s_M=\dfrac{s}{\sqrt{N}}=\dfrac{2.63}{\sqrt{4}}=\dfrac{2.63}{2}=1.315[/tex]
The degrees of freedom for this sample size are:
[tex]df=n-1=4-1=3[/tex]
The t-value for a 80% confidence interval and 3 degrees of freedom is t=1.638.
The margin of error (MOE) can be calculated as:
[tex]MOE=t\cdot s_M=1.638 \cdot 1.315=2.154[/tex]
Then, the lower and upper bounds of the confidence interval are:
[tex]LL=M-t \cdot s_M = 10.75-2.154=8.596\\\\UL=M+t \cdot s_M = 10.75+2.154=12.904[/tex]
The 80% confidence interval for the average net change is (8.596, 12.904).
Working on Summer Vacation. An Adweek/Harris (July 2011) poll found that 35% of U.S. adults do not work at all while on summer vacation. In a random sample of 10 U.S. adults, let x represent the number who do not work during summer vacation
a. For this experiment, define the event that represents a "success"
b. Explain why x is (approximately) a binomial random variable
c. Give the value of p for this binomial experiment
d. Find P(x = 3)
e. Find the probability that 2 or fewer of the 10 U.S. adults do not work during summer vacation
A becuase i worked it out and i got that so im really confident
A manager records the repair cost for 4 randomly selected stereos. A sample mean of $82.64 and standard deviation of $14.32 are subsequently computed. Determine the 90% confidence interval for the mean repair cost for the stereos. Assume the population is approximately normal. Step 1 of 2 : Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.
Answer:
CI = (70.861 , 94.418)
Step-by-step explanation:
In order to determine the 90% confidence interval you use the following formula (for a population approximately normal):
[tex]CI=(\overline{x}-Z_{\alpha/2}\frac{\sigma}{\sqrt{n}},\overline{x}+Z_{\alpha/2}\frac{\sigma}{\sqrt{n}})[/tex] (1)
[tex]\overline{x}[/tex]: mean = 82.64
σ: standard deviation = 14.32
n: sample = 4
α: tail area = 1 - 0.9 = 0.1
Z_α/2 = Z_0.05: Z factor = 1.645
You replace these values and you obtain:
[tex]Z_{0.05}(\frac{14.32}{\sqrt{4}})=(1.645)(\frac{14.32}{\sqrt{4}})=11.778[/tex]
The confidence interval will be:
[tex]CI=(82.64-11.778,82.64+11.778)=(70.861,94.418)[/tex]
The 90% confidence interval is (70.861 , 94.418)
Verona is solving the equation –3 + 4x = 9. In order to isolate the variable term using the subtraction property of equality, which number should she subtract from both sides of the equation? –4 –3 3 4
Answer:
subtract -3
Step-by-step explanation:
–3 + 4x = 9
Add 3 to each side
This is the same as subtracting -3
-3 + 4x - (-3) = 9 - (-3)
4x = 9 +3
4x = 12
What’s the correct answer for this ? Two chords AB and CD intersect at E. If AE = 2cm, EB =4, and CE = 2.5 cm, find the length of ED
Answer:
ED = 3.2 cm
Step-by-step explanation:
According to chord-chord power theorem,
(AE)(EB) = (CE)(ED)
2*4 = 2.5 *ED
8/2.5 = ED
ED = 3.2 cm
In a sample of seven cars, each car was tested for nitrogen-oxide emissions (in grams per mile) and the following results were obtained. 0.10 0.13 0.16 0.15 0.14 0.08 0.15 (a) Construct a 99% confidence interval for the mean nitrogen-oxide emissions of all cars. (b) If the EPA requires that nitrogen-oxide emissions be less than 0.165 g/mi, based on the 99% confidence interval in (a),
The question is incomplete. Here is the complete qeustion.
In a sample of seven cars, each car was tested for nitrogen-oxide emissions (in grams per mile) and the following results were obtained: 0.10 0.13 0.16 0.15 0.14 0.008 0.15
(a) Construct a 99% confidence interval for the mean nitrogen-oxide emissions of all cars.
(b) If the EPA requires that nitrogen-oxide emissions be less than 0.165 g/mi, based on the 99% confidence interval in (a), can we safely conclude that this requirement is being met?
Answer: (a) 0.089 ≤ μ ≤ 0.171
(b) No
Step-by-step explanation:
(a) To determine the confidence interval, first calculate the mean (X) and standard deviation (s) of the sample
X = [tex]\frac{0.1+0.13+0.16+0.15+0.14+0.08+0.15}{7}[/tex]
X = 0.13
s = [tex]\sqrt{\frac{(0.1-0.13)^{2} + (0.13 - 0.13)^{2} + ... + (0.15 - 0.13)^{2}}{7-1} }[/tex]
s = 0.029
The degrees of freedom is
N - 1 = 7 - 1 = 6
And since the confidence is of 99%:
α = 1 - 0.99 = 0.01
α/2 = 0.01/2 = 0.005
The t-test statistics for [tex]t_{6,0.005}[/tex] is 3.707
(Value found in the t-distribution table)
Now, calculate Error:
E = [tex]t_{6,0.005}[/tex] . [tex]\frac{s}{\sqrt{N} }[/tex]
E = 3.707. [tex]\frac{0.029}{\sqrt{7} }[/tex]
E = 0.041
The interval will be:
0.13 - 0.041 ≤ μ ≤ 0.13+0.041
0.089 ≤ μ ≤ 0.171
(b) No, because according to the interval, the nitrode-oxide emissions range from 0.089 to 0.171, which is greater than required by EPA.
Hitchhiker Snails A type of small snail is very widespread in Japan, and colonies of the snails that are genetically similar have been found very far apart. Scientists wondered how the snails could travel such long distances. A recent study1 provides the answer. Biologist Shinichiro Wada fed live snails to birds and found that of the snails were excreted live out the other end. The snails apparently are able to seal their shells shut to keep the digestive fluids from getting in.
What is the best estimate for the proportion of all snails of this type to live after being eaten by a bird?
Answer: 0.149
Step-by-step explanation:
As Scientists wondered how the snails could travel such long distances. A recent study provides the answer. Biologist Shinichiro Wada fed 174 live snails to birds and found that 26 of the snails were excreted live out the other end.
The best estimate for the proportion of all snails of this type to live after being eaten by a bird can be achieved by calculating the ratio of survival/number of eaten snails
Where the number of eaten snails = 174
The number of survivors = 26
Estimated proportion = 26/174 = 0.1494
Therefore, the best estimate for the proportion of all snails of this type to live after being eaten by a bird will be 0.149 approximately.
The base of a rectangular prism is 20 cm 2. If the volume of the prism is 100 cm 3, what is its height?
Answer:
Step-by-step explanation:
Answer:
height = 5
Step-by-step explanation:
The volume of a prism is V = l*w*h
You are not given any information about the exact values of l and w.
You do know however that L and w when multiplied together = 20, so you can put that in for l*w. Then the formula becomes
V = 20*h
You are told that the volume is 100. Now the problem is simplified. You get
100 = 20 * h Divide both sides by 20
100/20 = 20*h/20 Combine like terms.
5 = h
Noaya read a book cover to cover in a single session, at a rate of 55 pages per hour. After 4 hours, he had 350 pages left to read. Let y represent the number of pages left to read after x hours.
Answer: –55x + 570
Step-by-step explanation:
The person above me completely missed the question so this is the right one
Please answer this correctly
Answer:
the base is 5 meters long.
Step-by-step explanation:
To find the a missing side when given the area you have to multiply the area by 2 which in this case would be 50 so 50/10 would be 5m which is your answer.
So the right answer is 5m
Please see the attached picture for full solution
Hope it helps
Good luck on your assignment..
In how many ways can a president and a vice president be randomly selected from a class of 20 students?
Answer:
n how many ways can a president, vice president, and a secretary be chosen? It is 12X11X10. Permutation of n things taken r at a time: nPr=n!/(n-r)! 12P3=12*11*10*9!/9!= 12*11*10=1320 ways.
Step-by-step explanation:
|x+12| =-9
Pls help!!!!
Answer:
x=-21
Step-by-step explanation:
x+12=-9
minus twelve on both sides
-9-12 equals -21
x=-21
The yearbook club is handing out T-shirts to its members. There are 5 blue, 7 green, 9 red, and 4 yellow T-shirts in all. If Jacob is handed a T-shirt, what is the probability that the color is red? 925 916 35 1625
Answer:
9/25
Step-by-step explanation:
Don't worry, I just took this test and got it correct. Best of luck to you all!
we choose a sample of size 100 from a population of monthly cable bills having standard deviation $20 If we assume the population mean bill is $65, what is the probability mean of our sample is greater than $70. .
Answer:
0.62% probability that the mean of our sample is greater than $70.
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal probability distribution
When the distribution is normal, we use 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.
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
In this question, we have that:
[tex]\mu = 65, \sigma = 20, n = 100, s = \frac{20}{\sqrt{100}} = 2[/tex]
What is the probability mean of our sample is greater than $70.
This is 1 subtracted by the pvalue of Z when X = 70. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{70 - 65}{2}[/tex]
[tex]Z = 2.5[/tex]
[tex]Z = 2.5[/tex] has a pvalue of 0.9938
1 - 0.9938 = 0.0062
0.62% probability that the mean of our sample is greater than $70.
y
The figure shows A XYZ. XW is the angle
bisector of ZYXZ.
8
6.5
W
What is W Z?
Enter
your answer in the box. Do not round
your answer.
x
Z
6
units
Basic
Answer:
3.84 units
Step-by-step explanation:
By the properties of angle bisectors, ...
WZ/ZX = WY/YX
Solving for WY, we have ...
WY = (YX)(WZ)/(ZX) = (6.5/6)(WZ)
The length YZ is ...
YZ = 8 = WY +WZ
8 = (6.5/6)(WZ) +WZ = 12.5/6(WZ) . . . . substitute for WY
WZ = 8(6/12.5) . . . . multiply by 6/12.5
WZ = 3.84
Answer:
The correct answer is indeed 3.84 units
Step-by-step explanation:
I just took the test and got it correct hope this helps ☺
Which graph represents the function f(x) = |x| – 4? On a coordinate plane, an absolute value graph has a vertex at (0, 4). On a coordinate plane, an absolute value graph has a vertex at (negative 4, 0). On a coordinate plane, an absolute value graph has a vertex at (0, negative 4). On a coordinate plane, an absolute value graph has a vertex at (4, 0).
Answer:
(0, -4)
Step-by-step explanation:
The graph that represents the function is (c) on a coordinate plane, an absolute value graph has a vertex at (0, -4)
The equation of the function is given as:
[tex]f(x) = |x| - 4[/tex]
The above function is an absolute value function shifted down by 4 units
Hence, the graph that represents the function is a graph that has its vertex at (0,-4)
Read more about absolute value graphs at:
https://brainly.com/question/2166748
Antonio burns 75 calories for every 15 minutes
Answer is 5 calories/min
75 divided by 15 is 5
Answer:five cal per min.15 in five groups equals ''75''
What is the mixed number3 3/8 as a fraction
Answer: Mixed Number to Fraction
Mixed Numbers to Improper Fraction
Mixed Numbers Improper Fraction
3 3/4 15/4
3 3/8 27/8
3 8/9 35/9
Hope this helps!!
m is directly proportional to r squared when r=2 m=14 work out the value of r when m = 224
Answer:
32
Step-by-step explanation:
r:m
2:14
1:7
m=224
r=224 divided by 7
224/7=32
Edit: unless it is proportional to r^2 in which case it is a different answer
Answer:
m=504
Step-by-step explanation:
If your answer is in decimal form, round to 2 decimal places. For example, if your answer is .412, enter it into the fill-in-the-blank as .41, or if your answer is 1.415, enter it into the fill-in-the-blank as 1.42. If there are no decimals in your answer, you will simply enter the number; so if your answer is 2, enter 2 with no decimals. Do not enter any extra spaces, do not enter commas or $ or % symbols.
Answer:
If there is no decimal just put number and if the number is higher than 5 you round up example 1.46 you would put 1.5
Step-by-step explanation:
HOPE IT HELPS
n a random sample of 10 residents of the state of Florida, the mean waste recycled per person per day was 2.8 pounds with a standard deviation of 0.64 pounds. Determine the 95% confidence interval for the mean waste recycled per person per day for the population of Florida. Assume the population is approximately normal. Step 1 of 2 : Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places
Answer:
The critical value is T = 2.2622.
The 95% confidence interval for the mean waste recycled per person per day for the population of Florida is between 1.352 pounds and 4.248 pounds.
Step-by-step explanation:
We have the standard deviation for the sample, so we use the t-distribution to solve this question.
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 10 - 1 = 9
95% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 9 degrees of freedom(y-axis) and a confidence level of [tex]1 - \frac{1 - 0.95}{2} = 0.975[/tex]. So we have T = 2.2622, which is the critical value.
The margin of error is:
M = T*s = 2.2622*0.64 = 1.448
In which s is the standard deviation of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 2.8 - 1.448 = 1.352 pounds
The upper end of the interval is the sample mean added to M. So it is 2.8 + 1.448 = 4.248 pounds
The 95% confidence interval for the mean waste recycled per person per day for the population of Florida is between 1.352 pounds and 4.248 pounds.
A local coffee house surveyed 317 customers regarding their preference of chocolate chip or cranberry walnut scones . 150 customers prefer the Cranberry Walnut Scones . 81 customers who responded were males and prefer the Chocolate Chip Scones . 172 female customers responded . Find the probability that a customer chosen at random will be a male or prefer the Chocolate Chip Scones .
1. 25.6%
2. 24.1%
3. 72.9%
4. 98.4%
Answer:
3. 72.9%
Step-by-step explanation:
Let's call M the event that the customer is male and C the event that the customer prefer chocolate chips Scones.
So, the probability P(M∪C) that a customer chosen at random will be a male or prefer the Chocolate Chip Scones is calculated as:
P(M∪C) = P(M) + P(C) - P(M∩C)
Then, there are 145 males (317 customer - 172 females = 145 males), so the probability that the customer is a males is:
P(M) = 145/317 = 0.4574
There are 167 customers that prefer chocolate chips Scones ( 317 customers - 150 customers that prefer the Cranberry Walnut Scones = 167), so the probability that a customer prefer chocolate chips Scones is:
P(C) = 167/317 = 0.5268
Finally, 81 customers were males and prefer the Chocolate Chip Scones, so the probability that a customer will be a male and prefer chocolate chip scones is:
P(M∩C) = 81/317 = 0.2555
Therefore, P(M∪C) is equal to:
P(M∪C) = 0.4574 + 0.5268 - 0.2555
P(M∪C) = 0.7287
P(M∪C) = 72.9%
Answer:
3. 72.9%
Step-by-step explanation:
A probability is the number of desired outcomes divided by the number of total outcomes.
Desired outcomes:
Male or prefers the Chocolate Chip Scones. That is, males and females who prefer the Chocolate Chip Scones.
There are 172 female customers and 317-172 = 145 male customers.
150 customers prefer the Cranberry Walnut Scones. So 317 - 150 = 167 customers prefer the Chocolate Chip Scones.
81 of those are male, so 167 - 81 = 86 are female.
So the total of desired outcomes is 86 + 145 = 231
Total outcomes:
317 total customers.
Probability:
231/317 = 0.729
So the correct answer is:
3. 72.9%
What is the graph of 3x+5y=15
Answer: y= -15 - 3x/5 is the answer.
Step-by-step explanation:
Answer:
The second graph
Step-by-step explanation:
PLEASE HALP ME! ( WILL MARK BRAINLIEST! Thank you! ;)
Answer: 1/20
Step-by-step explanation:
Decimal Fraction Percentage
0.05 1/20 5%
Six measurements were made of the magnesium ion concentration (in parts per million, or ppm) in a city's municipal water supply, with the following results. It is reasonable to assume that the population is approximately normal. Based on a 95% confidence interval for the mean magnesium ion concentration, is it reasonable to believe that the mean magnesium ion concentration may be greater than 199.5? (Hint: you should first calculate the 95% confidence interval for the mean magnesium ion concentration.)
a) The likelihood cannot be determined
b) Yes
c) No
Answer:
Option B is correct.
It is reasonable to believe that the mean magnesium ion concentration may be greater than 199.5 as the confidence interval obtained contains values that are greater than 199.5
Step-by-step explanation:
Complete Question
Six measurements were made of the magnesium ion concentration (in parts per million, or ppm) in a city's municipal water supply, with the following results. It is reasonable to assume that the population is approximately normal.
170 201 199 202 173 153
Based on a 95% confidence interval for the mean magnesium ion concentration, is it reasonable to believe that the mean magnesium ion concentration may be greater than 199.5? (Hint: you should first calculate the 95% confidence interval for the mean magnesium ion concentration.)
A) The likelihood cannot be determined.
B) Yes
C) No
Solution
For this question, obtaining the confidence interval will give a clear solution to the problem.
Since the Confidence Interval for the population mean is basically an interval of range of values where the true population mean can be found with a certain level of confidence, if the range obtained contains values greater than the standard we are comparing against (199.5), then the confidence interval proves that the mean magnesium ion may be greater than 199.5.
But to obtain the confidence interval, we need the mean and standard deviation for the sample.
170, 201, 199, 202, 173, 153
Mean = (sum of variables)/(total number of variables)
Sum of variables = 170+201+199+202+173+153 = 1098
Total number of variables = 6
Mean = (1098/6) = 183
Standard deviation = σ = √[Σ(x - xbar)²/N]
x = each variable
xbar = mean = 183
N = number of variables = 6
Σ(x - xbar)² = (170-183)² + (201-183)² + (199-183)² + (202-183)² + (173-183)² + (153-183)²
= 169 + 324 + 256 + 361 + 100 + 900
= 2110
σ = √(2110/6) = 18.75
Mathematically,
Confidence Interval = (Sample mean) ± (Margin of error)
Sample Mean = 183
Margin of Error is the width of the confidence interval about the mean.
It is given mathematically as,
Margin of Error = (Critical value) × (standard Error of the mean)
Critical value will be obtained using the t-distribution. This is because there is no information provided for the population mean and standard deviation.
To find the critical value from the t-tables, we first find the degree of freedom and the significance level.
Degree of freedom = df = n - 1 = 6 - 1 = 5.
Significance level for 95% confidence interval
(100% - 95%)/2 = 2.5% = 0.025
t (0.025, 5) = 2.57 (from the t-tables)
Standard error of the mean = σₓ = (σ/√n)
σ = standard deviation of the sample = 18.75
n = sample size = 6
σₓ = (18.75/√6) = 7.656
95% Confidence Interval = (Sample mean) ± [(Critical value) × (standard Error of the mean)]
CI = 183 ± (2.57 × 7.656)
CI = 183 ± 19.675
95% CI = (163.325, 202.675)
95% Confidence interval = (163.3, 202.7)
It is reasonable to believe that the mean magnesium ion concentration may be greater than 199.5 as the confidence interval obtained contains values that are greater than 199.5
Hope this Helps!!!
help pls take your time..
Answer:
As [tex]{x \to \infty}, \,\,{f(x) \to -\infty[/tex] and as [tex]{x \to -\infty}, \,\,{f(x) \to \infty[/tex]
Step-by-step explanation:
Please look at the plotted points in the attached image. There we see that as x grows toward infinity (to the right), the values for f(x) seem to become more negative (so f(x) seems to go towards minus infinity).
As we move towards the left with values of x (x going towards negative infinity, f(x) seems to become more and more positive (grow toward infinity)
Once a bill leaves the Congress, how can a bill become a law? (Select all that anny
Answer:
Once both the House and Senate have approved the bill in identical form, it becomes "Enrolled" and sent to the President of the United States. The President may sign the bill into law. The President can also take no action on the bill for ten days while Congress is in session and the bill will automatically become law.
Suppose you are planning an experiment and a sample has yet been selected. For this experiment you plan on taking a SRS of 50 mice with pancreatic cancer measuring a particular hormone level. What would be the impact on a 95% confidence interval calculated from the experiment on these mice if instead of a SRS of 50 mice, a SRS of 200 mice were taken?
Answer:
The width or range of the confidence interval with sample size 200 will be about half of that of the confidence interval with sample 50.
Step-by-step explanation:
Confidence Interval for the population mean is basically an interval of range of values where the true population mean can be found with a certain level of confidence.
Mathematically,
Confidence Interval = (Sample mean) ± (Margin of error)
Margin of Error is the width of the confidence interval about the mean.
It is given mathematically as,
Margin of Error = (Critical value) × (standard Error of the mean)
Confidence Interval = (Sample mean) ± [(Critical value) × (standard Error of the mean)
- For the two random samples, of sizes 50 and 200, the Central limit theorem allows us to say that the sample mean is approximately equal to the population mean as this random sample satisfies the condition of being a simple random sample and a distribution obtained from a normal distribution.
- Making the right assumption that population standard deviation is known and z-distribution is used to find the critical value
Critical value for 95% = 1.96
The critical value for both samples are the same then.
- Standard Error of the mean = σₓ = (σ/√n)
where σ = population standard deviation
n = sample size
For the two distributions
Confidence Interval = (Sample mean) ± [(Critical value) × (Standard Error of the mean)
(Sample mean)₅₀ = (Sample mean)₂₀₀
(Critical value)₅₀ = (Critical value)₂₀₀
(Standard Error of the mean)₅₀ = (σ/√50) = 0.1414σ
(Standard Error of the mean)₂₀₀ = (σ/√200) = 0.0707σ
0.1414σ = 2 × 0.0707σ
(Standard Error of the mean)₅₀ = 2 × (Standard Error of the mean)₂₀₀
(Standard Error of the mean)₅₀ > (Standard Error of the mean)₂₀₀
Hence,
(Margin of Error)₅₀ > (Margin of Error)₂₀₀
(Margin of Error)₅₀ = 2 × (Margin of Error)₂₀₀
Confidence Interval = (Sample mean) ± (Margin of error)
Hence, the width or range of the confidence interval with sample size 50 will be about two times larger than the confidence interval with sample 200.
Hope this Helps!!!
A broker has calculated the expected values of two different financial instruments X and Y. Suppose that E(x)= $100, E(y)=$90 SD(x)= 90$ and SD(y)=$8. Find each of the following.
a. E(X+ 10) and SD(X+ 10)
b. E(5Y) and SD(5Y)
c) E(X+ Y) and SD(X+ Y)
d) What assumption must you make in part c?
Expectation is linear, meaning
E(a X + b Y) = E(a X) + E(b Y)
= a E(X) + b E(Y)
If X = 1 and Y = 0, we see that the expectation of a constant, E(a), is equal to the constant, a.
Use this property to compute the expectations:
E(X + 10) = E(X) + E(10) = $110
E(5Y) = 5 E(Y) = $450
E(X + Y) = E(X) + E(Y) = $190
Variance has a similar property:
V(a X + b Y) = V(a X) + V(b Y) + Cov(X, Y)
= a^2 V(X) + b^2 V(Y) + Cov(X, Y)
where "Cov" denotes covariance, defined by
E[(X - E(X))(Y - E(Y))] = E(X Y) - E(X) E(Y)
Without knowing the expectation of X Y, we can't determine the covariance and thus variance of the expression a X + b Y.
However, if X and Y are independent, then E(X Y) = E(X) E(Y), which makes the covariance vanish, so that
V(a X + b Y) = a^2 V(X) + b^2 V(Y)
and this is the assumption we have to make to find the standard deviations (which is the square root of the variance).
Also, variance is defined as
V(X) = E[(X - E(X))^2] = E(X^2) - E(X)^2
and it follows from this that, if X is a constant, say a, then
V(a) = E(a^2) - E(a)^2 = a^2 - a^2 = 0
Use this property, and the assumption of independence, to compute the variances, and hence the standard deviations:
V(X + 10) = V(X) ==> SD(X + 10) = SD(X) = $90
V(5Y) = 5^2 V(Y) = 25 V(Y) ==> SD(5Y) = 5 SD(Y) = $40
V(X + Y) = V(X) + V(Y) ==> SD(X + Y) = √[SD(X)^2 + SD(Y)^2] = √8164 ≈ $90.35
John averages 58 words per minute on a typing test with a standard deviation of 11 words per minute. Suppose John's words per minute on a minute on a typing test. Then X~N(58,11)
Answer:
The z score when x =72 is:
[tex] z = \frac{72-58}{11}= 1.273[/tex]
The mean is 58
This z-score tells you that x= 72 is 1.273 standard deviations to the right of the mearn.
Step-by-step explanation:
Assuming the following info for the question: Suppose John's words per minute on a typing test are normally distributed. Let X -the number of words per minute on a typing test. Then X N(58, 11) If necessary, round to three decimal places.
Provide your answer below rds per minute in a typing test on Sunday. The z score when x =72 is
For this case we know that the variable of interest is modelled with the normal distribution:
[tex]X \sim N (\mu= 58, \sigma=11)[/tex]
And the z score is given by:
[tex] z = \frac{X -\mu}{\sigma}[/tex]
And replacing we got:
[tex] z = \frac{72-58}{11}= 1.273[/tex]
The mean is 58
This z-score tells you that x= 72 is 1.273 standard deviations to the right of the mearn.
What preserves a shapes orientation?
a. Vertical translation
b. Reflection across the shapes base
c. Rotation about its center
Answer:
a.vertical translation
