Answer:
The probability that the mean daily precipitation will be of X inches or less for a random sample of n November days is the p-value of [tex]Z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex], in which [tex]\mu[/tex] is mean amount of inches of rain and [tex]\sigma[/tex] is the standard deviation.
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score 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 p-value, we get the probability that the value of the measure is greater than X.
Central Limit Theorem
The Central Limit Theorem establishes 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:
Mean [tex]\mu[/tex], standard deviation [tex]\sigma[/tex]
n days:
This means that [tex]s = \frac{\sigma}{\sqrt{n}}[/tex]
Applying the Central Limit Theorem to the z-score formula.
[tex]Z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
What is the probability that the mean daily precipitation will be of X inches or less for a random sample of November days?
The probability that the mean daily precipitation will be of X inches or less for a random sample of n November days is the p-value of [tex]Z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex], in which [tex]\mu[/tex] is mean amount of inches of rain and [tex]\sigma[/tex] is the standard deviation.
Suppose X has an exponential distribution with mean equal to 23. Determine the following:
(a) P(X >10)
(b) P(X >20)
(c) P(X <30)
(d) Find the value of x such that P(X
Answer:
a) P(X > 10) = 0.6473
b) P(X > 20) = 0.4190
c) P(X < 30) = 0.7288
d) x = 68.87
Step-by-step explanation:
Exponential distribution:
The exponential probability distribution, with mean m, is described by the following equation:
[tex]f(x) = \mu e^{-\mu x}[/tex]
In which [tex]\mu = \frac{1}{m}[/tex] is the decay parameter.
The probability that x is lower or equal to a is given by:
[tex]P(X \leq x) = \int\limits^a_0 {f(x)} \, dx[/tex]
Which has the following solution:
[tex]P(X \leq x) = 1 - e^{-\mu x}[/tex]
The probability of finding a value higher than x is:
[tex]P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}[/tex]
Mean equal to 23.
This means that [tex]m = 23, \mu = \frac{1}{23} = 0.0435[/tex]
(a) P(X >10)
[tex]P(X > 10) = e^{-0.0435*10} = 0.6473[/tex]
So
P(X > 10) = 0.6473
(b) P(X >20)
[tex]P(X > 20) = e^{-0.0435*20} = 0.4190[/tex]
So
P(X > 20) = 0.4190
(c) P(X <30)
[tex]P(X \leq 30) = 1 - e^{-0.0435*30} = 0.7288[/tex]
So
P(X < 30) = 0.7288
(d) Find the value of x such that P(X > x) = 0.05
So
[tex]P(X > x) = e^{-\mu x}[/tex]
[tex]0.05 = e^{-0.0435x}[/tex]
[tex]\ln{e^{-0.0435x}} = \ln{0.05}[/tex]
[tex]-0.0435x = \ln{0.05}[/tex]
[tex]x = -\frac{\ln{0.05}}{0.0435}[/tex]
[tex]x = 68.87[/tex]
Will mark Brainlest formula of (x+y)cube
Answer:
(x+y) ³
= x³+3x²y+3xy²+y³
=x³+y³+3xy(x+y)
Brainliest please~
(x+y)3= x^3+3x^2y+3xy^2+y^3
1. Rita is hiking along a trail that is 113.7 miles long. On the first day she hiked of 1 10 the distance of the trail. On the second day she hiked the same distance as the first day. How much of the trail does she have left to hike?
Answer:
She has 90.96 miles of the trail to hike.
Step-by-step explanation:
Length of the trail:
The length of the trail is of 113.7 miles.
On the first day she hiked of 1/10 the distance of the trail.
Thus:
[tex]\frac{1}{10(113.7) = 11.37[/tex]
On the first day she hiked 11.37 miles.
On the second day she hiked the same distance as the first day.
Also 11.37 miles on the second day, and thus, 2*11.37 = 22.74 miles on the first two days.
How much of the trail does she have left to hike?
113.7 - 22.74 = 90.96
She has 90.96 miles of the trail to hike.
Help me pls I don’t know how to do this
Answer:
[tex]radius=6.68cm[/tex]
Step-by-step explanation:
Formula to find radius:
[tex]r=\frac{C}{2\pi }[/tex]
[tex]r=42/2\pi[/tex]
[tex]r=42/2(3.14)[/tex]
[tex]r=6.68cm[/tex]
hope this helps......
Which expression is equivalent to the given expression?
6ab/(a^0b^2)^4
Answer:
,here is the answer
Step-by-step explanation:
here is your answer
the points -6/5 and -5/6 will line in which quadrant
A negative x is to the left of the y axis and a negative y value is below the x axis. Any value to the left and below the axis’ will be in the 3rd quadrant.
Answer: 3rd quadrant
25. Approximate the sample variance and standard deviation given the following frequency distribution: Class Frequency 0–9 13 10–19 7 20–29 10 30–39 9 40–49 11
Sample variance = 228.408
Standard deviation = 15.113
Step-by-step explanation:The well formatted frequency table has been attached to this response.
To calculate the sample variance and standard deviation of the given grouped data, follow these steps:
i. Find the midpoint (m) of the class interval.
This is done by adding the lower bounds and upper bounds of the class intervals and dividing the result by 2. i.e
For class 0 - 9, we have
m = (0 + 9) / 2 = 4.5
For class 10 - 19, we have
m = (10 + 19) / 2 = 14.5
For class 20 - 29, we have
m = (20 + 29) / 2 = 24.5
For class 30 - 39, we have
m = (30 + 39) / 2 = 34.5
For class 40 - 49, we have
m = (40 + 49) / 2 = 44.5
This is shown in the third column of the attached table.
ii. Find the product of each of the frequencies of the class intervals and their corresponding midpoints. i.e
For class 0 - 9, we have
frequency (f) = 13
midpoint (m) = 4.5
=> f x m = 13 x 4.5 = 58.5
For class 10 - 19, we have
frequency (f) = 7
midpoint (m) = 14.5
=> f x m = 7 x 14.5 = 101.5
For class 20 - 29, we have
frequency (f) = 10
midpoint (m) = 24.5
=> f x m = 10 x 24.5 = 245
For class 30 - 39, we have
frequency (f) = 9
midpoint (m) = 34.5
=> f x m = 9 x 34.5 = 310.5
For class 40 - 49, we have
frequency (f) = 11
midpoint (m) = 44.5
=> f x m = 11 x 44.5 = 489.5
This is shown in the fourth column of the attached table.
iii. Calculate the mean (x) of the distribution i.e
This is done by finding the sum of all the results in (ii) above and dividing the outcome by the sum of the frequencies. i.e
x = ∑(f x m) ÷ ∑f
Where;
∑(f x m) = 58.5 + 101.5 + 245 + 310.5 + 489.5 = 1205
∑f = 13 + 7 + 10 + 9 + 11 = 50
=> x = 1205 ÷ 50
=> x = 24.1
Therefore, the mean is 24.1
This is shown on the fifth column of the attached table.
iv. Calculate the deviation of the midpoints from the mean.
This is done by finding the difference between the midpoints and the mean. i.e m - x where x = mean = 24.1 and m = midpoint
For class 0 - 9, we have
midpoint (m) = 4.5
=> m - x = 4.5 - 24.1 = -19.6
For class 10 - 19, we have
midpoint (m) = 14.5
=> m - x = 14.5 - 24.1 = -9.6
For class 20 - 29, we have
midpoint (m) = 24.5
=> m - x = 24.5 - 24.1 = 0.4
For class 30 - 39, we have
midpoint (m) = 34.5
=> m - x = 34.5 - 24.1 = 10.4
For class 40 - 49, we have
midpoint (m) = 44.5
=> m - x = 44.5 - 24.1 = 20.4
This is shown on the sixth column of the attached table.
v. Find the square of each of the results in (iv) above.
This is done by finding (m-x)²
For class 0 - 9, we have
=> (m - x)² = (-19.6)² = 384.16
For class 10 - 19, we have
=> (m - x)² = (-9.6)² = 92.16
For class 20 - 29, we have
=> (m - x)² = (0.4)² = 0.16
For class 30 - 39, we have
=> (m - x)² = (10.4)² = 108.16
For class 40 - 49, we have
=> (m - x)² = (20.4)² = 416.16
This is shown on the seventh column of the attached table.
vi. Multiply each of the results in (v) above by their corresponding frequencies.
This is done by finding f(m-x)²
For class 0 - 9, we have
=> f(m - x)² = 13 x 384.16 = 4994.08
For class 10 - 19, we have
=> f(m - x)² = 7 x 92.16 = 645.12
For class 20 - 29, we have
=> f(m - x)² = 10 x 0.16 = 1.6
For class 30 - 39, we have
=> f(m - x)² = 9 x 108.16 = 973.44
For class 40 - 49, we have
=> f(m - x)² = 11 x 416.16 = 4577.76
This is shown on the eighth column of the attached table.
vi. Calculate the sample variance.
Variance σ², is calculated by using the following relation;
σ² = ∑f(m-x)² ÷ (∑f - 1)
This means the variance is found by finding the sum of the results in (vi) above and then dividing the result by one less than the sum of all the frequencies.
∑f(m-x)² = sum of the results in (vi)
∑f(m-x)² = 4994.08 + 645.12 + 1.6 + 973.44 + 4577.76 = 11192
∑f - 1 = 50 - 1 = 49 {Remember that ∑f was calculated in (iii) above}
∴ σ² = 11192 ÷ 49 = 228.408
Therefore, the variance is 228.408
vii. Calculate the standard deviation
Standard deviation σ, is calculated by using the following relation;
σ =√ [ ∑f(m-x)² ÷ (∑f - 1) ]
This is done by taking the square root of the variance calculated above.
σ = [tex]\sqrt{228.408}[/tex]
σ = 15.113
Therefore, the standard deviation is 15.113
Suppose 2500 dollars is invested at 4% for 5 years. Find the account balance if it is
compounded daily, monthly, quarterly?
answer step by step
Answer:
Kindly check explanation
Step-by-step explanation:
Using the compound interest formula :
A = P(1 + r/n)^nt
A = final amount ; r = rate ; n = number of compounding times per period ; t = period ; P = principal
P = 2500 ; r = 4% = 0.04 ; t = 5 years
Daily compounding, n = 365
Yearly compounding, n = 1
Quarterly compounding, n = 4
Daily compounding :
A = 2500(1 + 0.04/365)^(5*365)
A = 2500(1.0001095)^1825
A = $3053.4734
Yearly :
A = 2500(1 + 0.04/1)^(5*1)
A = 2500(1.04)^5
A = $3041.6323
Quarterly:
A = 2500(1 + 0.04/4)^(5*4)
A = 2500(1.01)^20
A = $3050.4751
Factor completely, then place the factors in the proper location on the grid. a8 - 12a4 + 36
Answer:
[tex]{ \tt{ {a}^{8} - {12a}^{4} + 36}} \\ = { \tt{ {a}^{4} ( {a}^{2} - 12) + 36 }} \\ = ( {a}^{2} - 12)( {a}^{4} + 36) \\ [/tex]
A professor has been teaching introductory statistics for many years and the final exam performance has been very consistent from class to class and the scores have been normally distributed. Overall, the whole data base (i.e. population) of final scores has a mean (μ) of 24 points (out of a maximum of 30 points) and a standard deviation (Ï) of 5 points. The professor would like to revise the course design to see if student performance on the final could be improved.
The new course design was implemented in the most recent academic year. There were 100 students and the average final exam score was 24.7. The professor would like to run a hypothesis test to see if this sample of students in the recent academic year performed significantly better than the past population. In other words, the hypothesis was a comparison between the population taking the course with the new design (represented by the sample of 100 students) with the population taking the course with the old design. The professor is predicting an increase of final score with the new design, so the hypotheses should be directional, and the test should be one-tailed. The significance level is set at α = .1.
Required:
a. Identify the dependent variable for this study.
b. State the null hypothesis and alternative hypothesis using both words and symbol notation
Answer:
a) Independent variable - Design of the course
Dependent variable - Final score of the students
b) H0 - Final score >24.7
Alternate hypothesis - Final score is less than or equal to 24.7
Step-by-step explanation:
a) Independent variable - Design of the course
Dependent variable - Final score of the students
b) Null Hypothesis : Performance of student taking course with the new design is better as compared to the population of student taking the course with the old design.
H0 - Final score >24.7
Alternate hypothesis - Final score is less than or equal to 24.7
Suppose your marketing colleague used a known population mean and standard deviation to compute the standard error as 67.5 for samples of a particular size. You don't know the particular sample size but your colleague told you that the sample size is greater than 70. Your boss asks what the standard error would be if you quadruple (4x) the sample size. What is the standard error for the new sample size
Answer:
The standard error for the new sample size will be of 33.75.
Step-by-step explanation:
Central Limit Theorem
The Central Limit Theorem establishes 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.
Standard error as 67.5 for samples of a particular size.
We have that [tex]s = \frac{\sigma}{\sqrt{n}}[/tex], that is, the standard error is inversely proportional to the square root of the sample size, so if you quadruple (4x) the sample size, the standard error will be divided by half. So
67.5/2 = 33.75
The standard error for the new sample size will be of 33.75.
Solve for T: 10t-4x=3S Explanation plz
solve for x 6(x-3)=8(x-4)
Answer:
7=x
Step-by-step explanation:
6(x-3)=8(x-4)
Distribute
6x -18 = 8x-32
Subtract 6x from each side
6x-18 -6x = 8x-32-8x
-18 = 2x-32
Add 32 to each side
-18+32 = 2x-32+32
14 = 2x
Divide by 2
14/2 =2x/2
7=x
[tex]\sf\purple{x= 7}[/tex]
[tex]\sf \bf {\boxed {\mathbb {STEP-BY-STEP\:\:EXPLANATION:}}}[/tex]
[tex]➺\:6(x - 3) = 8(x - 4)[/tex]
[tex]➺ \: 6x - 18 = 8x - 32[/tex]
[tex]➺ \: 6x - 8x = - 32 + 18[/tex]
[tex]➺ \: - 2x = - 14[/tex]
[tex]➺ \: x = \frac{ - 14}{ - 2} [/tex]
[tex]➺ \: x = 7[/tex]
Therefore, the value of [tex]x[/tex] is 7.
[tex]\sf \bf {\boxed {\mathbb {TO\:VERIFY :}}}[/tex]
[tex]➺ \: 6(x - 3) = 8(x - 4)[/tex]
[tex]➺ \: 6(7 - 3) = 8(7 - 4)[/tex]
[tex]➺ \: 6 \times 4 = 8 \times 3[/tex]
[tex]➺ \: 24 = 24[/tex]
➺ L. H. S. = R. H. S.
Hence verified.
[tex]\red{\large\qquad \qquad \underline{ \pmb{{ \mathbb{ \maltese \: \: Mystique35ヅ}}}}}[/tex]
What is the smallest number you should subtract from 456 to make it divisible by 9?
Joseph borrows $10000 from his sister Katie at an annual interest rate of 10%. If the
interest is compounded twice a year, how much does he owe after 12 months? Give your answer in dollars.
Answer:
A = P ( 1 + r / n) ^( t * n)
where
A = the amt owed
P = amt borrowed
r = the interest rate as a decimal
n = the number of compoundings per year
t = the number of years
A = 10000 ( 1 + .10 / 2)^(2 *1) = 10000 ( 1.05)^2 = $11025
Step-by-step explanation:
pleaseeeee helpppppppppppp
9514 1404 393
Answer:
maximum height: 26.5 ftair time: 2.5 secondsStep-by-step explanation:
I find the easiest way to answer these questions is to use a graphing calculator. It can show you the extreme values and the intercepts. The graph below shows the maximum height is 26.5 ft. The time in air is about 2.5 seconds.
__
If you prefer to solve this algebraically, you can use the equation of the axis of symmetry to find the time of the maximum height:
t = -b/(2a) = -(40)/(2×-16) = 5/4
Then the maximum height is ...
h(5/4) = -16(5/4)² +40(5/4) +1.5 = -25 +50 +1.5 = 26.5 . . . feet
__
Now that we know the vertex of the function, we can write it in vertex form:
h(t) = -16(t -5/4)² +26.5
Solving for the value of t that makes this zero, we get ...
0 = -16(t -5/4)² +26.5
16(t -5/4)² = 26.5
(t -5/4)² = 26.5/16 = 1.65625
Then ...
t = 1.25 +√1.65625 ≈ 2.536954
The cannon ball is in the air about 2.5 seconds.
simplification please
Answer:
5
Step-by-step explanation:
WHen we raise a power to a power, we multiply them, in this case 5 is the base so we can just ignore it for now and replace it with x.
(X^1/3)^3
Multiply 1/3 by 3 and we get 1
So:
X^1
Which does nothing, so we can simplify to just:\
X
Remember x is 5 so the answer is:
5
Write down in terms of n, an expression for the nth term
of the following sequences:
a) 6 2 -2 -6 -10
b) -8 -15 -22 -29 -36
Answer:
[tex]{ \bf{(a).}} \\ { \tt{ {n}^{th} = a + (n - 1)d }} \\ { \tt{ {n}^{th} = 6 + (n - 1) \times - 4 }} \\ {n}^{th} = 10 - 4n \\ \\ { \bf{(b).}} \\ { \tt{ {n}^{th} = - 8 + (n- 1) \times - 7 }} \\ { \tt{ {n}^{th} = -1 - 7n}}[/tex]
Find each missing length to the nearest tenth.
[tex]\huge\bold{Given:}[/tex]
Length of the perpendicular = 7
Length of the base = 10
[tex]\huge\bold{To\:find:}[/tex]
The length of the missing side (hypotenuse).
[tex]\large\mathfrak{{\pmb{\underline{\orange{Solution}}{\orange{:}}}}}[/tex]
[tex]\longrightarrow{\purple{x\:=\: 12.21}}[/tex]
[tex]\large\mathfrak{{\pmb{\underline{\red{Step-by-step\:explanation}}{\red{:}}}}}[/tex]
Let the length of the missing side be [tex]x[/tex].
Using Pythagoras theorem, we have
(Hypotenuse)² = (Perpendicular)² + (Base)²
[tex]\longrightarrow{\blue{}}[/tex] [tex]{x}^{2}[/tex] = (7)² + (10)²
[tex]\longrightarrow{\blue{}}[/tex] [tex]{x}^{2}[/tex] = 49 + 100
[tex]\longrightarrow{\blue{}}[/tex] [tex]{x}^{2}[/tex] = 149
[tex]\longrightarrow{\blue{}}[/tex] [tex]x[/tex] = [tex]\sqrt{149}[/tex]
[tex]\longrightarrow{\blue{}}[/tex] [tex]x[/tex] = 12.206
[tex]\longrightarrow{\blue{}}[/tex] [tex]x[/tex] = 12.21.
Therefore, the length of the missing side [tex]x[/tex] is [tex]12.21[/tex].
[tex]\huge\bold{To\:verify :}[/tex]
[tex]\longrightarrow{\green{}}[/tex] (12.21)² = (7)² + (10)²
[tex]\longrightarrow{\green{}}[/tex] 149 = 49 + 100
[tex]\longrightarrow{\green{}}[/tex] 149 = 149
[tex]\longrightarrow{\green{}}[/tex] L.H.S. = R. H. S.
Hence verified.
[tex]\huge{\textbf{\textsf{{\orange{My}}{\blue{st}}{\pink{iq}}{\purple{ue}}{\red{35}}{\green{ヅ}}}}}[/tex]
6
Which expression is equivalent
Answer:
I thimk it is B
Step-by-step explanation:
Solve for x round to the nearest tenth if necessary
Answer:
x = 2.8
Step-by-step explanation:
sin = opp/hyp
hyp = opp/sin
x = 2.4/sin60
x = 2.771281292110204
rounded
x = 2.8
Question 11 of 40
Factor this polynomial completely.
x2 - 6x + 9
A. (x+3)(x+3)
B. Does not factor
C. (x-3)(x - 3)
D. (x+3)(x-3)
Anyone plz show how to work it out step by step.
Answer:
168cm^3
Step-by-step explanation:
Q to P is going to be 3cm. it is identical to the length T to U.
R to T , W to Q, S to U is going to be identical to P to V. P to V has been identified as 12 cm.
in the middle of the shape, there are 4 identical triangles. the height time length will give us the area of that one shape:
e.g for shape P to V to W to Q and back to P is one rectangle. the length is 12 cm and the width is 3 cm.
12 x 3= 36
36cm^3 is one rectangles surface area, we have 4 identical triangles that means we need to times 36 by 4.
so 36x4=144.
now on the left and right side, we have two squares. on the right, we have T to U to V to W back to T this has the height of 3 width of 4 then we do 3 X 4 which is 12, we times it by 2 because we have two identical squares.
12 X 2=24
finally we add 24 and 144 = 168cm^3.
hope this helps :)
The perimeter of a parallelogram must be no less than 40 feet. The length of the rectangle is 6 feet. What are the possible measurements of the width? Write an inequality to represent this problem. Use w to represent the width of the parallelogram. [Hint: The formula for finding the perimeter of a parallelogram is P = 2 l + 2 w . What is the smallest possible measurement of the width? Justify your answer by showing all your work.
Answer: [tex]14\ ft[/tex]
Step-by-step explanation:
Given
Length of rectangle is [tex]6\ ft[/tex]
Perimeter must be greater than 40 ft
Suppose l and w be the length and width of the rectangle
[tex]\Rightarrow \text{Perimeter P=}2(l+w)\\\Rightarrow P\geq 40\\\Rightarrow 2(l+w)\geq40\\\Rightarrow l+w\geq20\\\Rightarrow w\geq20-6\\\Rightarrow w\geq14\ ft[/tex]
So, the smallest width can be [tex]14\ ft[/tex]
Last month Rudy’s Tacos sold 22 dinner specials. The next month they released a new commercial and sold 250% of last month’s dinners. How many dinner specials did they sell this month?
Answer:
the answer is 2
Step-by-step explanation: because 250 -22 is i dont even know
Answer:
55
Step-by-step explanation:
What is this function’s input if its output is 11?
f(x) = 2x + 5
Answer:
the input x is 3
Step-by-step explanation:
2x+5=11
2x=6
x=3
Find the 23rd term of the arithmetic sequence with the terms a1 27 and d = 16.
Answer:
379
Step-by-step explanation:
a23 = 27 + (23-1)(16)
= 27 + (22)(16)
= 27 + 352
= 379
Find the interest earned on $1,000 for 1 year at a 6% rate of interest when the interest is compounded quarterly.
Answer:
1060
Step-by-step explanation:
A minority representation group accuses a major bank of racial discrimination in its recent hires for financial analysts. Exactly 16% of all applications were from minority members, and exactly 15% of the 2100 open positions were filled by members of the minority.
Required:
a. Find the mean of p, where p is the proportion of minority member applications in a random sample of 2100 that is drawn from all applications.
b. Find the standard deviation of p.
Answer:
a) The mean is of [tex]\mu = 0.16[/tex]
b) The standard deviation is of [tex]s = 0.008[/tex]
Step-by-step explanation:
Central Limit Theorem
The Central Limit Theorem establishes 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.
For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean [tex]\mu = p[/tex] and standard deviation [tex]s = \sqrt{\frac{p(1-p)}{n}}[/tex]
Question a:
Exactly 16% of all applications were from minority members
This means [tex]p = 0.16[/tex], and thus, the mean is of [tex]\mu = p = 0.16[/tex]
b. Find the standard deviation of p.
2100 open positions, thus [tex]n = 2100[/tex].
[tex]s = \sqrt{\frac{p(1-p)}{n}}[/tex]
[tex]s = \sqrt{\frac{0.16*0.84}{2100}}[/tex]
[tex]s = 0.008[/tex]
The standard deviation is of [tex]s = 0.008[/tex]
Find the value of each shape so that they will add up to give you the specified sums in each row and each column.
How does this solving problem relate to system of equations?
Answer:
Step-by-step explanation:
