Answer:
76 ≤ B ≤ 89
Step-by-step explanation:
Mean score (μ) = 73.3
Standard deviation (σ) = 9.7
If B Scores are below the top 5% and above the bottom 62%, then:
62% ≤ B ≤ 95%
In a normal distribution, the 62nd percentile has a corresponding z-score of z = 0.305, while the 95th percentile has a corresponding z-score of z = 1.650.
The grades X1 and X2 which are the limits for a B grade are given by:
[tex]z= \frac{X-\mu}{\sigma}\\0.305= \frac{X_1-73.3}{9,7}\\X_1=76.2\\1.650= \frac{X_2-73.3}{9,7}\\X_2=89.3[/tex]
Rounding to the nearest whole number, a B grade is given to grades between 76 and 89.
Final answer:
To calculate the numerical limits for a B grade in a normally distributed set of test scores with a mean of 73.3 and a standard deviation of 9.7, we find the z-scores corresponding to the top 5% and bottom 62%. The B grade range is approximately between 70 and 89.
Explanation:
To find the numerical limits for a B grade in a normally distributed set of test scores, we must determine the z-scores that correspond to the top 5% and the bottom 38% of scores since a B grade is assigned to scores below the top 5% and above the bottom 62% (which is equivalent to being below the top 38%). Given the mean score is 73.3 and the standard deviation is 9.7, we can use z-score formulas to convert these percentages to raw scores.
For the top 5% cutoff (z = 1.645), the score is calculated as follows:
Top 5% cutoff score = mean + (z x standard deviation)
Top 5% cutoff score = 73.3 + (1.645 x 9.7)
Top 5% cutoff score = 73.3 + 15.9565
Top 5% cutoff score ≈ 89
For the top 38% cutoff (z = -0.31), which is the bottom 62%, the score is calculated as:
Bottom 62% cutoff score = mean + (z x standard deviation)
Bottom 62% cutoff score = 73.3 + (-0.31 x 9.7)
Bottom 62% cutoff score = 73.3 + (-2.997)
Bottom 62% cutoff score ≈ 70
Therefore, a B grade on this test would be for scores roughly between 70 and 89.
Montoya (2007) asked 56 men and 82 women to rate 21 different body parts on a scale of 1 (no opinion) to 5 (very desirable). They found that men and women rated the eyes similarly, with an average rating of about 3.77 ± 1.23 (M ± SD). Assuming these data are normally distributed, answer the following questions. (Round your answers to two decimal places.) (a) What percentage of participants rated the eyes at least a 5 (very desirable)? % (b) What percentage rated the eyes at most a 1 (no opinion)? %
Answer:
a. 15.9%
b. 1.2%
Step-by-step explanation:
using the normal distribution we have the following expression
[tex]P(X\geq a)=P(\frac{X-u}{\alpha } \geq \frac{5-M}{SD}) \\[/tex]
Where the first expression in the right hand side is the z-scores, M is the mean of value 3.77 and SD is the standard deviation of value 1.23.
if we simplify and substitute values, we arrive at
[tex]P(X\geq a)=P(\frac{X-u}{\alpha } \geq \frac{a-M}{SD}) \\P(X\geq 5)=P(Z \geq \frac{5-3.77}{1.23})\\P(X\geq 5)=P(Z \geq 1)\\P(X\geq 5)=1- P(Z < 5)\\P(X\geq 5)=1-0.841\\P(X\geq 5)=0.159[/tex]
in percentage, we arrive at 15.9%
b. for the percentage rated the eyes most a 1
[tex]P(X\leq a)=P(\frac{X-u}{\alpha } \leq \frac{a-M}{SD})\\for a=1\\P(X\leq 1)=P(Z \leq \frac{1-3.77}{1.23})\\P(X\leq 1)=P(Z \leq -2.25})\\P(X\leq 1)=0.012\\[/tex]
In percentage we have 1.2%
The numerical course grades in a statistics course can be approximated by a normal model with a mean of 70 and a standard deviation of 10. The professor must convert the numerical grades to letter grades. She decides that she wants 10% A's, 30% B's, 40% C's, 15% D's, and 5% F's. a. What is the cutoff for an A grade?
Answer:
The cutoff for an A grade is 82.8.
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 = 70, \sigma = 10[/tex]
a. What is the cutoff for an A grade?
The top 10% of the class get an A grade. So the cutoff is the value of X when Z has a pvalue of 1-0.1 = 0.9. So it is X when Z = 1.28
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]1.28 = \frac{X - 70}{10}[/tex]
[tex]X - 70 = 1.28*10[/tex]
[tex]X = 82.8[/tex]
The cutoff for an A grade is 82.8.
solution to 2/3x = -5/2 + 2
Answer:
x=-0.75
Step-by-step explanation:
Combine Like terms
2/3x= -5/2+2
2/3x= -2.5+2
2/3x= -0.5
Multiply both sides by 3
2x= -1.5
Divide both sides by 2
x= -0.75
Answer:
x = -4/3
Step-by-step explanation:
The LCM for the fractions is 6x.
2/3x = -5/2 + 2
Multiply through by 6x
(6x) 2/3x = (6x) -5/2 + (6x)2
4 = -15x + 12x
4 = -3x
x = -4/3
Enjoy Maths!
Seventy independent messages are sent from an electronic transmission center.Messages are processed sequentially, one after another. Transmission time of each message is Exponential with parameter λ = 5 min−1. Find the probability that all 70 messages are transmitted in less than 12 minutes. Use the Central Limit Theorem.
Answer:
The probability that all 70 messages are transmitted in less than 12 minutes is 0.117.
Step-by-step explanation:
Let X = the transmission time of each message.
The random variable X follows an Exponential distribution with parameter λ = 5 minutes.
The expected value of X is:
[tex]E(X)=\frac{1}{\lambda}=\frac{1}{5}=0.20[/tex]
The variance of X is:
[tex]V(X)=\frac{1}{\lambda^{2}}=\frac{1}{5^{2}}=0.04[/tex]
Now define a random variable T as:
T = X₁ + X₂ + ... + X₇₀
According to a Central limit theorem if a large sample (n > 30) is selected from a population with mean μ and variance σ² then the sum of random variables X follows a Normal distribution with mean, [tex]\mu_{s} = n\mu[/tex] and variance, [tex]\sigma^{2}_{s}=n\sigma^{2}[/tex].
Compute the probability of T < 12 as follows:
[tex]P(T<12)=P(\frac{T-\mu_{T}}{\sqrt{\sigma_{T}^{2}}}<\frac{12-(70\times0.20)}{\sqrt{70\times 0.04}})\\=P(Z<-1.19)\\\=1-P(Z<1.19)\\=1-0.883\\=0.117[/tex]
*Use a z-table for the probability.
Thus, the probability that all 70 messages are transmitted in less than 12 minutes is 0.117.
Following are the solution to the given question:
Let X signify the letter's transmission time [tex]i^{th}[/tex] , and [tex]i =1,2,3,...,70.[/tex] this is assumed that the [tex]X_i \sim Exp (\lambda =5)[/tex].
[tex]\to Mean =\frac{1}{\lambda} =\frac{1}{5}=0.2\\\\ \to Variance =\frac{1}{\lambda^2} =\frac{1}{5^2} =\frac{1}{25}=0.04\\\\[/tex]
Let [tex]T = X_1 + X_2 +...+ X_{70}[/tex], be the total transmission time. According to the Central Limit Theorem, T has a Normal distribution of mean, Variance.mean [tex]= E(T) = 70 \times 0.2=14[/tex]
Variance [tex]=V(T) =70 \times 0.04 = 2.8[/tex]
Therefore
[tex]\to P(T<12) = P (Z < \frac{12-14}{\sqrt{2.8}})[/tex]
[tex]= P(Z<-1.195) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (by sy memetry) \\\\=0.5-P(0<Z<1.19)\\\\=0.5-0.3830 \\\\=0.1170\\\\[/tex]
Therefore, the final answer is "0.1170".
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A certain college graduate borrows 7864 dollars to buy a car. The lender charges interest at an annual rate of 13%. Assuming that interest is compounded continuously and that the borrower makes payments continuously at a constant annual rate k dollars per year, determine the payment rate that is required to pay off the loan in 3 years. Also determine how much interest is paid during the 3-year period.
Answer:
Therefore rate of payment = $ 3145.72
Therefore the rate of interest = =$1573.17
Step-by-step explanation:
Consider A represent the balance at time t.
A(0)=$ 7864.
r=13 % =0.13
Rate payment = $k
The balance rate increases by interest (product of interest rate and current balance) and payment rate.
[tex]\frac{dB}{dt} = rB-k[/tex]
[tex]\Rightarrow \frac{dB}{dt} - rB=-k[/tex].......(1)
To solve the equation ,we have to find out the integrating factor.
Here p(t)= the coefficient of B =-r
The integrating factor [tex]=e^{\int p(t) dt[/tex]
[tex]=e^{\int (-r)dt[/tex]
[tex]=e^{-rt}[/tex]
Multiplying the integrating factor the both sides of equation (1)
[tex]e^{-rt}\frac{dB}{dt} -e^{-rt}rB=-ke^{-rt}[/tex]
[tex]\Rightarrow e^{-rt}dB - e^{-rt}rBdt=-ke^{-rt}dt[/tex]
Integrating both sides
[tex]\Rightarrow \int e^{-rt}dB -\int e^{-rt}rBdt=\int-ke^{-rt}dt[/tex]
[tex]\Rightarrow e^{-rt}B=\frac{-ke^{-rt}}{-r} +C[/tex] [ where C arbitrary constant]
[tex]\Rightarrow B(t)=\frac{k}{r} +Ce^{rt}[/tex]
Initial condition B=7864 when t =0
[tex]\therefore 7864= \frac{k}{r} - Ce^0[/tex]
[tex]\Rightarrow C= \frac{k}{r} -7864[/tex]
Then the general solution is
[tex]B(t)=\frac{k}{r}-( \frac{k}{r}-7864)e^{rt}[/tex]
To determine the payment rate, we have to put the value of B(3), r and t in the general solution.
Here B(3)=0, r=0.13 and t=3
[tex]B(3)=0=\frac{k}{0.13}-( \frac{k}{0.13}-7864)e^{0.13\times 3}[/tex]
[tex]\Rightarrow- 0.48\frac{k}{0.13} +11614.98=0[/tex]
⇒k≈3145.72
Therefore rate of payment = $ 3145.72
Therefore the rate of interest = ${(3145.72×3)-7864}
=$1573.17
The payment rate that is required to pay off the loan in 3 years is approximately 2949.51 dollars/year. The interest paid during the 3-year period is about 984.53 dollars.
Explanation:To solve this problem, the formula for continuously compounded interest should be used which is A = P * e^(rt), where A is the value of the investment at a future time t, P is the principal amount, r is the annual interest rate, and t is the time the money is invested or borrowed for.
First, set up the equation: 7864 = k * (1/(0.13)) * (e^(0.13 * 3) - 1), then solve for k: k = 7864 * (0.13) / (e^(0.13 * 3) - 1) ≈ 2949.51 dollars/year.
Using the formula A = P * e^(rt), the total amount paid is A = 2949.51 * 3 = 8848.53 dollars. The interest paid during the 3-year period is calculated by subtracting the loan amount from the total amount paid, that is 8848.53 - 7864 = 984.53 dollars.
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In a certain small dorm, the dorm council will consist of 8 students, ofwhom 3 must be women and 5 must be men, because there are a total of 15 women and 25 men currently living there. What is the total possible number of complete councils that could be selected
Answer:
The total possible number of complete councils that could be selected is 24,174,150.
Step-by-step explanation:
The order of the men and the women is not important. So we use the combinations formula to solve this question.
Combinations formula:
[tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
What is the total possible number of complete councils that could be selected
3 women from a set of 15
5 men from a set of 25
[tex]T = C_{15,3}*C_{25,5} = \frac{15!}{3!12!}*\frac{25!}{5!20!} = 455*53130 = 24174150[/tex]
The total possible number of complete councils that could be selected is 24,174,150.
marco and paolo are working with expressions with rational exponents. marco claims that 642/3=512.which of them is correct? use the rational exponent to justify your answer
Answer:
poalo is correct
Step-by-step explanation
Answer:
Neither Marco nor Paolo is correct, because neither one used the rational exponent property correctly.
Step-by-step explanation:
I just did the test and it told me that I got it right.
Hope that helps! Please mark me brainliest.
The daily sales S (in thousands of dollars) that are attributed to an advertising campaign are given by S = 11 + 3 t + 3 − 18 (t + 3)2 where t is the number of weeks the campaign runs. (a) Find the rate of change of sales at any time t.
Answer:
The parking rate at the hospital a parking garage are 50 cents for the 1st hour and 25 cents for each additional hour if guan part in the hospital parking garage for 8 hours how much will the total of charge for parking be
A builder of houses needs to order some supplies that have a waiting time Y for delivery, with a continuous uniform distribution over the interval from 1 to 4 days. Because she can get by without them for 2 days, the cost of the delay is fixed at $400 for any waiting time up to 2 days. After 2 days, however, the cost of the delay is $400 plus $50 per day (prorated) for each additional day. That is, if the waiting time is 3.5 days, the cost of the delay is $400 $50(1.5)
Find the expected value of the builder’s cost due to waiting for supplies.
Answer:
The Expected cosy of the builder is $433.3
Step-by-step explanation:
$400 is the fixed cost due to delay.
Given Y ~ U(1,4).
Calculating the Variable Cost, V
V = $0 if Y≤ 2
V = 50(Y-2) if Y > 2
This can be summarised to
V = 50 max(0,Y)
Cost = 400 + 50 max(0, Y-2)
Expected Value is then calculated by;
Waiting day =2
Additional day = at least 1
Total = 3
E(max,{0, Y - 2}) = integral of Max {0, y - 2} * ⅓ Lower bound = 1; Upper bound = 4, (4,1)
Reducing the integration to lowest term
E(max,{0, Y - 2}) = integral of (y - 2) * ⅓ dy Lower bound = 2; Upper bound = 4 (4,2)
E(max,{0, Y - 2}) = integral of (y) * ⅓ dy Lower bound = 0; Upper bound = 2 (2,0)
Integrating, we have
y²/6 (2,0)
= (2²-0²)/6
= 4/6 = ⅔
Cost = 400 + 50 max(0, Y-2)
Cost = 400 + 50 * ⅔
Cost = 400 + 33.3
Cost = 433.3
Answer:
the expected value of the builder’s cost due to waiting for supplies is $433.3
Step-by-step explanation:
Due to the integration symbol and also for ease of understanding, i have attached the explanation as an attachment.
Is there more wood in a 60-foot-high tree trunk with a radius of 2.2 feet or in a 50-foot-high tree trunk with a radius of 2.6 feet? Assume that the trees can be regarded as right circular cylinders.
Answer:
V₂=1061.85 ft³
Step-by-step explanation:
To determine where more wood is found, just find the volume of each log and see which one has the largest volume
knowing that the volume of a straight cylinder is
V=πR²h, where R=radius and h=height
[tex]V_{1}=\pi (2.2)^{2}60= 912.31ft^{3}\\V_{2}=\pi (2.6)^{2}50= 1061.85ft^{3}[/tex]
As we can observe
V₂>V₁
Assume lim f(x)-6 and lim g(x)-9. Compute the following limit and state the limit laws used to justify the computation.
x→3 x →3
Lim 3√f(x).g(x)+10
x →3
Using the limit laws of addition and multiplication, the limit of the given equation is solved by combining the calculated limits of the function and the constant, giving an answer of approximately 13.78.
Explanation:To solve this problem, we can use the limit laws of addition and multiplication. The limit laws state that the limit of the sum of two functions is equal to the sum of their individual limits, and the limit of the product of two functions is equal to the product of their individual limits.
So, based on these laws, we first separate the function into two parts: the limit of 3√(f(x).g(x)) as x approaches 3 and the limit of 10 as x approaches 3.
Given lim f(x)=>6 and lim g(x)=>9 when x approaches 3, we multiply these values together to get a product of 54. The cubed root of 54 is approximately 3.78.
The constant 10 has a limit of 10, as constants maintain their value irrespective of the limit.
Adding these values together, 3.78 + 10, gives us a limit of approximately 13.78.
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5. (20 pts) Consider the array 25, 14, 63, 29, 63, 47, 12, 21. Apply the Split procedure in Quicksort, as described in class, to this array using the first element as the pivot.
Answer:
Step-by-step explanation: see attachment
Suppose that grade point averages of undergraduate students at one university have a bell-shaped distribution with a mean of 2.52 and a standard deviation of 0.38. Using the empirical rule, what percentage of the students have grade point averages that are between 1.76 and 3.28
Using the empirical rule for a bell-shaped distribution, which states that nearly all data for a normal distribution falls within three standard deviations of the mean, approximately 95% of undergraduate students at this university have grade point averages between 1.76 and 3.28.
Explanation:The student is asking for the percentage of GPA between 1.76 and 3.28 using the Empirical Rule which applies to a Bell-shaped distribution or normal distribution. The empirical rule states that for a normal distribution, nearly all of the data will fall within three standard deviations of the mean.
The mean in this case is 2.52 and the standard deviation is 0.38. Therefore the values 1.76 and 3.28 fall within the mean minus two standard deviations and mean plus two standard deviations respectively. Therefore, by the empirical rule, these values represent approximately 95% of the data.
In other words, according to the empirical rule, approximately 95% of undergraduate students at this university have grade point averages between 1.76 and 3.28.
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Consider the following data on x = rainfall volume (m3) and y = runoff volume (m3) for a particular location. x 8 12 14 19 23 30 40 45 55 67 72 79 96 112 127 y 4 10 13 14 15 25 27 44 38 46 53 74 82 99 105 Use the general software output to decide whether there is a useful linear relationship between rainfall and runoff.
There is a useful linear relationship between rainfall and runoff and the relationship is y = 0.85x - 2.65
Deciding whether there is a useful linear relationship between rainfall and runoff
From the question, we have the following parameters that can be used in our computation:
x 8 12 14 19 23 30 40 45 55 67 72 79 96 112 127
y 4 10 13 14 15 25 27 44 38 46 53 74 82 99 105
Using a graphing tool, we have the following summary
Sum of X = 799Sum of Y = 649Mean X = 53.2667Mean Y = 43.2667Sum of squares (SSX) = 20086.9333Sum of products (SP) = 17313.9333The regression equation can be represented as
y = bx + a
Where
b = SP/SSX = 17313.93/20086.93 = 0.86
a = MY - bMX = 43.27 - (0.86*53.27) = -2.65
So, we have
y = 0.85x - 2.65
Hence, there is a useful linear relationship between rainfall and runoff
Suppose the null hypothesis, H0, is: a sporting goods store claims that at least 70% of its customers do not shop at any other sporting goods stores. What is the Type I error in this scenario? a. The sporting goods store thinks that less than 70% of its customers do not shop at any other sporting goods stores when, in fact, less than 70% of its customers do not shop at any other sporting goods stores. b. The sporting goods store thinks that at least 70% of its customers do not shop at any other sporting goods stores when, in fact, at least 70% of its customers do not shop at any other sporting goods stores. c. The sporting goods store thinks that less than 70% of its customers do not shop at any other sporting goods stores when, in fact, at least 70% of its customers do not shop at any other sporting goods stores. d. The sporting goods store thinks that at least 70% of its customers do not shop at any other sporting goods stores when, in fact, less than 70% of its customers do not shop at any other sporting goods stores.
Answer:
Null hypothesis: [tex]p \geq 0.7[/tex]
Alternative hypothesis: [tex]p<0.7[/tex]
A type of error I for this case would be reject the null hypothesis that the population proportion is greater or equal than 0.7 when actually is not true.
So the correct option for this case would be:
c. The sporting goods store thinks that less than 70% of its customers do not shop at any other sporting goods stores when, in fact, at least 70% of its customers do not shop at any other sporting goods stores.
Step-by-step explanation:
Previous concepts
A hypothesis is defined as "a speculation or theory based on insufficient evidence that lends itself to further testing and experimentation. With further testing, a hypothesis can usually be proven true or false".
The null hypothesis is defined as "a hypothesis that says there is no statistical significance between the two variables in the hypothesis. It is the hypothesis that the researcher is trying to disprove".
The alternative hypothesis is "just the inverse, or opposite, of the null hypothesis. It is the hypothesis that researcher is trying to prove".
Type I error, also known as a “false positive” is the error of rejecting a null hypothesis when it is actually true. Can be interpreted as the error of no reject an alternative hypothesis when the results can be attributed not to the reality.
Type II error, also known as a "false negative" is the error of not rejecting a null hypothesis when the alternative hypothesis is the true. Can be interpreted as the error of failing to accept an alternative hypothesis when we don't have enough statistical power.
Solution to the problem
On this case we want to test if the sporting goods store claims that at least 70^ of its customers, so the system of hypothesis would be:
Null hypothesis: [tex]p \geq 0.7[/tex]
Alternative hypothesis: [tex]p<0.7[/tex]
A type of error I for this case would be reject the null hypothesis that the population proportion is greater or equal than 0.7 when actually is not true.
So the correct option for this case would be:
c. The sporting goods store thinks that less than 70% of its customers do not shop at any other sporting goods stores when, in fact, at least 70% of its customers do not shop at any other sporting goods stores.
In a new card game, you start with a well-shuffled full deck and draw 3 cards without replacement. If you draw 3 hearts, you win $50. If you draw 3 black cards, you win $25. For any other draws, you win nothing.
(a) Find the expected winnings for a single game.
(b) Find the standard deviation of the winnings.
(c) If the game costs $5 to play, what would be the expected value of the net profit (or loss)?
(Hint: profit = winnings - cost; X - 5)
(d) If the game costs $5 to play, what would be the standard deviation of the net profit (or loss)?
(e) If the game costs $5 to play, should you play this game?
There are [tex]\binom{52}3=\frac{52!}{3!(52-3)!}[/tex] (or "52 choose 3") ways of drawing any 3 cards from the deck.
There are 13 hearts in the deck, and 26 cards with a black suit. So there are [tex]\binom{13}3[/tex] and [tex]\binom{26}3[/tex] ways of drawing 3 hearts or 3 black cards, respectively. Then the probability of drawing 3 hearts is
[tex]P(\text{3 hearts})=\dfrac{\binom{13}3}{\binom{52}3}=\dfrac{11}{850}[/tex]
and the probability of drawing 3 black cards is
[tex]P(\text{3 black})=\dfrac{\binom{26}3}{\binom{52}3}=\dfrac2{17}[/tex]
All other combinations can be drawn with probability [tex]1-\frac{11}{850}-\frac2{17}=\frac{739}{850}[/tex].
Let [tex]W[/tex] be the random variable for one's potential winnings from playing the game. Then
[tex]P(W=w)=\begin{cases}\frac{11}{850}&\text{for }w=\$50\\\frac2{17}&\text{for }w=\$25\\\frac{739}{850}&\text{otherwise}\end{cases}[/tex]
a. For a single game, one can expect to win
[tex]E[W]=\displaystyle\sum_ww\,P(W=w)=\frac{\$50\cdot11}{850}+\frac{\$20\cdot2}{17}+\frac{\$0\cdot739}{850}=\$3[/tex]
b. For a single game, one's winnings have a variance of
[tex]V[W]=E[(W-E[W])^2]=E[W^2]-E[W]^2[/tex]
where
[tex]E[W^2]=\displaystyle\sum_ww^2\,P(W=w)=\frac{\$50^2\cdot11}{850}+\frac{\$20^2\cdot2}{17}+\frac{\$0^2\cdot739}{850}=\$^2\frac{1350}{17}\approx\$^279.41[/tex]
so that [tex]V[W]=\$^2\frac{1197}{17}\approx\$^270.41[/tex]. (No, that's not a typo, variance is measured in squared units.) Standard deviation is equal to the square root of the variance, so it is approximately $8.39.
c. With a $5 buy-in, the expected value of the game would be
[tex]E[W-\$5]=E[W]-\$5=-\$2[/tex]
i.e. a player can expect to lose $2 by playing the game (on average).
d. With the $5 cost, the variance of the winnings is the same, since the variance of a constant is 0:
[tex]V[W-\$5]=V[W][/tex]
so the standard deviation is the same, roughly $8.39.
e. You shouldn't play this game because of the negative expected winnings. The odds are not in your favor.
The expected winning for a single game defined is : $3.59
The standard deviation of winning is : $10.11
Expected winning if game costs $5 to play is :
- $0.79
The standard deviation of winning if game costs $5 to play is : $11.33
The game should not be played with a game play fee of -$5 as the expected winning value is negative.
Recall : selection is done without replacement :
Number of hearts in a deck = 13
Probability of drawing 3 hearts :
P(drawing 3 Hearts)
First draw × second draw × third draw
13/52 × 12/51 × 11/50 = 1716/132600 = 858/66300
Probability of selecting 3 black cards :
Number of black cards in a deck = 26
P(drawing 3 black cards) :
First draw × second draw × third draw
26/52 × 25/51 × 24/50 = 15600 / 132600 = 7800/66300
Probability of making any other draw :
P(3 hearts) + P(3 blacks) + P(any other draw) = 1
858/66300 + 7800/66300 + P(any other draw) = 1
P(any other draw) = 57642/66300
For a single game :
X _______ $50 ________ $25 _______ $0
P(X)_ 858/66300__ 7800/66300_57642/66300
E(X) = Σ[ X × p(X)]
E(X) =Σ[(50 × 858/66300)+(25 × 7800/66300)+0]
E(X) = $3.588
The standard deviation = √Var(X)
Var(X) = Σ[ X² × p(X)] - E(X)
Σ[ X²×p(X)] = Σ[(50² × 858/66300)+(25² × 7800/66300)+0] = 105.88235
Var(X) = 105.88235 - 3.588 = 102.29435
Standard deviation of winning = √102.29435 = $10.114
If the game cost $5 to play :
Net amount won if :
3 hearts are drawn = $50 - $5 = $45
3 blacks are drawn = $25 - $5 = $20
Any other combination are drawn= $0 - $5 = -$5
The distribution becomes :
X _______ $45 ________ $20 _______ -$5
P(X)_ 858/66300__ 7800/66300_57642/66300
E(X) = Σ[ X × p(X)]
E(X) =Σ[(50 × 858/66300)+(25 × 7800/66300)+ (-5 × 57642/66300)]
E(X) = - $0.7588
Standard deviation of winning :
The standard deviation = √Var(X)
Var(X) = Σ[ X² × p(X)] - E(X)
Σ[ X²×p(X)] = Σ[(50² × 858/66300)+(25² × 7800/66300)+ (-5² × 57642/66300)] = 127.61764
Var(X) = 127.61764 - (-0.7588) = 128.37644
Standard deviation of winning :
Std(X) = √Var(X) = √128.37644 = $11.330
With a game cost of - $5 ; the expected winning for a single game gives a negative value, therefore you should not play the game.
Learn more on expected value : https://brainly.com/question/22097128
If mABC= 184°, what is m∠ABC?
88°
90°
84°
92°
mADC=360-mABC=360-184=176⁰
So, measure of angle ABC= mADC/2=176/2=88⁰
The required measure of angle ABC is 88° for the given figure. The correct answer is option A.
What is an arc?The arc is a portion of the circumference of a circle. The circumference of a circle is the distance or perimeter around a circle
The measure of the arc is given as follows:
mABC = 184°
According to the given figure, mADC is the part of the full circle which is complete arc that measure of angle is 360°.
mADC = 360° - mABC
mADC = 360° - 184°
mADC = 176°
As we know that the measure of angle ABC is equal to half of mADC.
The measure of angle ABC = mADC/2
The measure of angle ABC = 176/2
The measure of angle ABC = 88°
Therefore, the required measure of angle ABC is 88°.
Learn more about the arc of the circle here:
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A box contains five slips of paper, marked $1, $1, $10, $25, and $25. The winner of a contest selects two slips of paper at random and then gets the larger of the dollar amounts on the two slips. Define a random variable w by w = amount awarded. Determine the probability distribution of w. (Hint: Think of the slips as numbered 1, 2, 3, 4, and 5, so that an outcome of the experiment consists of two of these numbers.)
Answer:
w = $1 or $10 or $25
Probability distribution of w = ($1,$1 $1,$1 $1,$10 $1,$25 $1,$25 $1,$1 $1,$1 $1,$10 $1,$25 $1,$25 $10,$1 $10,$1 $10,$10 $10,$25 $10,$25 $25,$1 $25,$1 $25,$10 $25,$25 $25,$25 $25,$1 $25,$1 $25,$10 $25,$25 $25,$25)
Step-by-step explanation:
Since the winner gets the larger amount of the two slips picked, random variable 'w' which is the amount awarded could be $1 or $10 or $25.
The probability distribution of 'w' is the values that the statistic takes on. Which could be: $1,$1 $1,$1 $1,$10 $1,$25 $1,$25 $1,$1 $1,$1 $1,$10 $1,$25 $1,$25 $10,$1 $10,$1 $10,$10 $10,$25 $10,$25 $25,$1 $25,$1 $25,$10 $25,$25 $25,$25 $25,$1 $25,$1 $25,$10 $25,$25 $25,$25
Solve using normalcdf
Answer:
0.209
Step-by-step explanation:
Find the sample mean and standard deviation.
μ = 1050
s = 218 / √50 = 30.8
Find the z-score.
z = (1075 − 1050) / 30.8
z = 0.81
Find the probability.
P(Z > 0.81) = 1 − 0.7910 = 0.2090
Scarborough Faire Herb Farm is a small company specializing in selling organic fresh herbs, teas and herbal crafts. Currently, basil is their top selling herb, with $45,000 in sales last year. Parsley is their second biggest seller with $27,000 in sales. Total sales last year were $170,000 and Scarborough Faire forecasts sales to increase by 10% this year. If basil sales remain the same as last year but total sales grow as percentage will basil sales be this year?
Answer:
24.06% of total sales
Step-by-step explanation:
Total sales, which were $170,000 originally, are expected to grow by 10%. The expected value of total sales this year is:
[tex]S=\$170,000*1.10 = \$187,000[/tex]
If Basil sales remain at the same value of $45,000, the percentage of sales corresponding to basil is:
[tex]B=\frac{\$45,000}{\$187,000}*100\%\\B=24.06\%[/tex]
Therefore, basil will correspond to 24.06% of total sales.
Basil sales will represent approximately 24.06% of Scarborough Faire Herb Farm's projected total sales this year, given that total sales are forecasted to increase by 10% and basil sales remain unchanged.
Explanation:The student's question concerns the percentage of total sales that basil sales will represent for Scarborough Faire Herb Farm in the current year, assuming a 10% increase in total sales from the previous year and unchanged basil sales. Firstly, we calculate the projected total sales for this year by increasing last year's total sales by 10%. The calculation is as follows:
Total sales last year: $170,000Forecasted increase: 10%Projected total sales this year: $170,000 + ($170,000 × 0.10) = $170,000 + $17,000 = $187,000Since the basil sales are to remain the same as last year ($45,000), we now calculate what percentage this represents of the projected total sales for this year:
Basil sales this year: $45,000 (unchanged)Percentage of basil sales out of total sales: ($45,000 / $187,000) × 100%Final percentage: 24.06% (rounded to two decimal places)Thus, basil sales will account for approximately 24.06% of the Scarborough Faire Herb Farm's total sales this year.
Identify the correct statements for the given functions of set {a, b, c, d} to itself.
1.
f(a) = b, f(b) = a, f(c) = c, f(d) = d is a one-to-one function as each element is an image of exactly one element.
2.
f(a) = b, f(b) = a, f(c) = c, f(d) = d is not a one-to-one function as a is mapped to b and b is mapped to a.
3.
f(a) = b, f(b) = b, f(c) = d, f(d) = c is a one-to-one function as each element is mapped to only one element.
4.
f(a) = b, f(b) = b, f(c) = d, f(d) = c is not a one-to-one function as b is an image of more than one elements.
5.
f(a) = d, f(b) = b, f(c) = c, f(d) = d is a one-to-one function as each element is mapped to only one element.
6.
f(a) = d, f(b) = b, f(c) = c, f(d) = d is not a one-to-one function as d is an image of more than one element.
Answer:
The correct answers are
option(1), option (4),option (5)
Step-by-step explanation:
One to one: Every image has exactly one unique pre-image in domain.
(1)
f(a) = b, f(b)=a, f(c)=c, f(d)=d
b has a pre-image in domain i.e a
a has a pre-image in domain i.e b
c has a pre-image in domain i.e c
d has a pre-image in domain i.e d
Here all elements have a unique pre- image in domain.
Therefore it is one to one.
(2)
f(a) = b, f(b)=a, f(c)=c, f(d)=d
b has a pre-image in domain i.e a
a has a pre-image in domain i.e b
c has a pre-image in domain i.e c
d has a pre-image in domain i.e d
Here all elements have a unique pre- image in domain.
Therefore it is one to one.
(3)
f(a) = b, f(b)=b, f(c)=d, f(d)=c
b has a pre-image in domain i.e a
b has a pre-image in domain i.e b
d has a pre-image in domain i.e c
c has a pre-image in domain i.e d
b has two pre image.
Here all elements have not a unique pre- image in domain.
Therefore it is not a one to one mapping.
(4)
f(a) = b, f(b)=b, f(c)=d, f(d)=c
b has a pre-image in domain i.e a
b has a pre-image in domain i.e b
d has a pre-image in domain i.e c
c has a pre-image in domain i.e d
b has two pre image.
Here all elements have not a unique pre- image in domain.
Therefore it is not a one to one mapping.
(5)
f(a) = d, f(b)=b, f(c)=d, f(d)=c
d has a pre-image in domain i.e a
b has a pre-image in domain i.e b
d has a pre-image in domain i.e c
c has a pre-image in domain i.e d
Here all elements have a unique pre- image in domain.
Therefore it is a one to one mapping.
(6)
f(a)=d, f(b)=b,f(c)=c,f(d)=d
d has a pre-image in domain i.e a
b has a pre-image in domain i.e b
c has a pre-image in domain i.e c
d has a pre-image in domain i.e d
Here all elements have a unique pre- image in domain.
Therefore it is a one to one mapping.
f(a) = b, f(b) = a, f(c) = c, f(d) = d and f(a) = d, f(b) = b, f(c) = c, f(d) = d are one-to-one function and f(a) = b, f(b) = b, f(c) = d, f(d) = c is not one-to-one function. Then the correct statements are 1, 4, and 5.
What is a function?The function is an expression, rule, or law that defines the relationship between one variable to another variable. Functions are ubiquitous in mathematics and are essential for formulating physical relationships.
One-to-one - every image has exactly one unique pre-image in the domain.
1. f(a) = b, f(b) = a, f(c) = c, f(d) = d
b has a pre image in a domian that is a
a has a pre image in a domian that is b
c has a pre image in a domian that is c
d has a pre image in a domian that is d
Here all elements have a unique pre-image in a domain.
Therefore it is one-to-one.
2. f(a) = b, f(b) = a, f(c) = c, f(d) = d
b has a pre image in a domian that is a
a has a pre image in a domian that is b
c has a pre image in a domian that is c
d has a pre image in a domian that is d
Here all elements have a unique pre-image in a domain.
Therefore it is one-to-one.
3. f(a) = b, f(b) = b, f(c) = d, f(d) = c
b has a pre image in a domian that is a
b has a pre image in a domian that is b
d has a pre image in a domian that is c
c has a pre image in a domian that is d
Here all elements do have not a unique pre-image in a domain.
Therefore it is not one-to-one.
4. f(a) = b, f(b) = b, f(c) = d, f(d) = c
b has a pre image in a domian that is a
b has a pre image in a domian that is b
d has a pre image in a domian that is c
c has a pre image in a domian that is d
Here all elements do have not a unique pre-image in a domain.
Therefore it is not one-to-one.
5. f(a) = d, f(b) = b, f(c) = c, f(d) = d
d has a pre image in a domian that is a
b has a pre image in a domian that is b
c has a pre image in a domian that is c
d has a pre image in a domian that is d
Here all elements have a unique pre-image in a domain.
Therefore it is one-to-one.
6. f(a) = d, f(b) = b, f(c) = c, f(d) = d
d has a pre image in a domian that is a
b has a pre image in a domian that is b
c has a pre image in a domian that is c
d has a pre image in a domian that is d
Here all elements have a unique pre-image in a domain.
Therefore it is one-to-one.
More about the function link is given below.
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Jeremy and Brenda drove their cars in
opposite directions. When they stopped
after some time, they were already
126 miles apart. If Jeremy drove twice
as far as Brenda, how many miles did
Jeremy drive?
Answer: Jeremy drove 84 miles.
Step-by-step explanation:
Let x represent the number of miles that Brenda drove.
If Jeremy drove twice
as far as Brenda, it means that the distance covered by Jeremy would be 2x miles
When they stopped after some time, they were already
126 miles apart. This means that the total distance covered by both of them is 126 miles. Therefore,
x + 2x = 126
3x = 126
x = 126/3
x = 42 miles
The number of miles that Jeremy drove is
42 × 2 = 84 miles
A college senior who took the Graduate Record Examination exam scored 530 on the Verbal Reasoning section and 600 on the Quantitative Reasoning section. The mean score for Verbal Reasoning section was 460 with a standard deviation of 105, and the mean score for the Quantitative Reasoning was 429 with a standard deviation of 148. Suppose that both distributions are nearly normal. Round calculated answers to 4 decimal places unless directed otherwise.
Answer:
Step-by-step explanation:
Given that a college senior who took the Graduate Record Examination exam scored 530 on the Verbal Reasoning section and 600 on the Quantitative Reasoning section.
Quantitative reasons scores are N (429, 148)
Hence Z score for the college senior = [tex]\frac{530-429}{148} \\=0.6824[/tex]
The mean score for Verbal Reasoning section was 460 with a standard deviation of 105
Verbal reasoning score was N(460, 105)
The score of college senior = 530
Z score for verbal reasoning =[tex]\frac{530-460}{105} \\=0.6667[/tex]
comparing we can say he scored more on quantiative reasoning.
Thus comparison was possible only by converting to Z scores.
The accounting records of Blossom Company show the following data. Beginning inventory 3,090 units at $5 Purchases 7,930 units at $8 Sales 9,390 units at $10 Determine cost of goods sold during the period under a periodic inventory system using the FIFO method. (Round answer to 0 decimal places, e.g. 1,250.)
Answer:
The cost of goods sold is 65,850
Step-by-step explanation:
FIFO Perpetual chart is attached.
FIFO Perpetual chart shows purchases, sales and balance of the period.
The cost of goods sold is:
3,090 units x $5=$15,450
6,300 units x $8=$50,400
Total=65,850
Which of the following is equivalent to 60 Superscript one-half? StartFraction 60 Over 2 EndFraction StartRoot 60 EndRoot StartFraction 1 Over 60 squared EndFraction StartFraction 1 Over StartRoot 60 EndRoot EndFraction
Answer:
[tex](60)^{\frac{1}{2}} = \sqrt{60}[/tex]
Step-by-step explanation:
We are given the following expression:
[tex](60)^{\frac{1}{2}}[/tex]
We are given the following options:
[tex]A.~\dfrac{60}{2}\\\\B.~\sqrt{60}\\\\C.~(\dfrac{1}{60})^{2}\\\\D.~\dfrac{1}{\sqrt{60}}[/tex]
Exponent Properties:
[tex]x^{-a} = (\dfrac{1}{x})^{a}\\\\(x^m)^n = x^{mn}\\\\\dfrac{x^m}{x^n} = x^{m-n}[/tex]
Thus, the correct answer is
[tex](60)^{\frac{1}{2}} = \sqrt{60}[/tex]
Answer:
B or [tex]\sqrt{60}[/tex]
Step-by-step explanation:
An average light bulb manufactured at The Lightbulb Company lasts and average of 300 days, with a standard deviation of 50 days. Suppose the lifespan of a light bulb from this company is normally distributed. (a) What is the probability that a light bulb from this company lasts less than 210 days? More than 330 days?(b) What is the probability that a light bulb from this company lasts between 280 and 380 days?(c) How would you characterize the lifespan of the light bulbs whose lifespans are among the shortest 2% of all bulbs made by this company?(d) If a pack of 6 light bulbs from this company are purchased, what is the probability that exactly 4 of them last more than 330 days?
Answer:
a). 0.03593, 0.27425
(bl 0.0703
(c) the lifespan of such bulb is less than 210 days
(d) 0.0298.
Step-by-step explanation:
Given that U = 300, sd= 50
Z = X-U/sd
(a) We find the probability
Pr(X<210) = Pr(Z<(210-300)/50)
= Pr(Z<-1.8)
= 0.03593
Pr(X >330) = Pr(Z >(330-300)/50))
= Pr(Z> 0.6)
= 1- PR(Z<0.6) = 1 - 0.72575
Pr(X>330)= 0.27425
(b) Pr(280< X< 330) = Pr( 280 < Z< 380)
= Pr([280-300/50] < Z < [380-300/50])
= Pr(0.4 < Z < 1.6)
= Pr(Z< 1.6) - Pr(Z < 0.4)
=0.72575 - 0.65542
= 0.07033
= 0.0703
(c) the lifespan of such bulbs is less than 210 days
(d) given n = 6, x = 4 ,
Pr(X>330) = 0.27425 = p
q = 1-p = 0.72575
Pr(X=4) = 6C4 × (0.27425)⁴ ×(0.72575)²
Pr(X=4) = 10 × 0.005657 × 0.52671
Pr(X= 4) = 0.029795
Pr(X= 4) = 0.0298
Children inherit many traits from their parents. Some inherited traits are not desirable. For example, if both parents are carriers, there is a 25% chance any child will have sickle-cell disease. If we assume no identical twins in a family, and both individuals in a couple are carriers, what is the probability none of their four children has the disease
The probability that none of their four children has the disease = 0.316
Step-by-step explanation:
Step 1 :
The probability that the child will have the specified disease if both the parents are carriers = 25% = 0.25
So the probability that the child does not have the specified disease = 1 - 0.25 = 0.75
Step 2 :
There are four children. The probability that each child has the disease is independent of the condition of the other children
The probability of more than one independent events happening together can be obtained by multiplying probability of individual events
So, the probability that none of their 4 children has the sickle cell disease = [tex](0.75)^{4}[/tex] = 0.316
Step 3 :
Answer :
The probability no children has the specified cell disease = 0.316
Which product is positive?
Apply the properties of angles to solve for the missing angles. Angle y is what degrees. Angle x is what degrees.
Answer:
y= 65 degrees. and x=30 degrees
Step-by-step explanation:
i haven't done this for a while so i may or may not be correct i tried to help though :)
Answer: y = 52°. x = 64°
Step-by-step explanation:
y = 26 + 26 = 52 degrees ( exterior angle of a triangle is the sum of the two opposite interior angle of the triangle)
x = 180-(52+64)(sum of angles in a triangle is 180°)
x = 180-116
x = 64degrees ( base angles of an isosceles triangle are equal)
In an article regarding interracial dating and marriage recently appeared in a newspaper. Of 1719 randomly selected adults, 311 identified themselves as Latinos, 322 identified themselves as blacks, 251 identified themselves as Asians, and 775 identified themselves as whites. Among Asians, 79% would welcome a white person into their families, 71% would welcome a Latino, and 66% would welcome a black person.
NOTE: If you are using a Student's t-distribution, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.)
Construct the 95% confidence intervals for the three Asian responses.
1. Welcome a white person ( , )
2. Welcome a Latino ( , )
3. Welcome a Black person ( , )
Answer:
Step-by-step explanation:
Hello!
The parameter of interest in this exercise is the population proportion of Asians that would welcome a person of other races in their family. Using the race of the welcomed one as categorizer we can define 3 variables:
X₁: Number of Asians that would welcome a white person into their families.
X₂: Number of Asians that would welcome a Latino person into their families.
X₃: Number of Asians that would welcome a black person into their families.
Now since we are working with the population that identifies as "Asians" the sample size will be: n= 251
Since the sample size is large enough (n≥30) you can apply the Central Limit Theorem and approximate the variable distribution to normal.
[tex]Z_{1-\alpha /2}= Z_{0.975}= 1.965[/tex]
1. 95% CI for Asians that would welcome a white person.
If 79% would welcome a white person, then the expected value is:
E(X)= n*p= 251*0.79= 198.29
And the Standard deviation is:
V(X)= n*p*(1-p)= 251*0.79*0.21=41.6409
√V(X)= 6.45
You can construct the interval as:
E(X)±Z₁₋α/₂*√V(X)
198.29±1.965*6.45
[185.62;210.96]
With a 95% confidence level, you'd expect that the interval [185.62; 210.96] contains the number of Asian people that would welcome a White person in their family.
2. 95% CI for Asians that would welcome a Latino person.
If 71% would welcome a Latino person, then the expected value is:
E(X)= n*p= 251*0.71= 178.21
And the Standard deviation is:
V(X)= n*p*(1-p)= 251*0.71*0.29= 51.6809
√V(X)= 7.19
You can construct the interval as:
E(X)±Z₁₋α/₂*√V(X)
178.21±1.965*7.19
[164.08; 192.34]
With a 95% confidence level, you'd expect that the interval [164.08; 192.34] contains the number of Asian people that would welcome a Latino person in their family.
3. 95% CI for Asians that would welcome a Black person.
If 66% would welcome a Black person, then the expected value is:
E(X)= n*p= 251*0.66= 165.66
And the Standard deviation is:
V(X)= n*p*(1-p)= 251*0.66*0.34= 56.3244
√V(X)= 7.50
You can construct the interval as:
E(X)±Z₁₋α/₂*√V(X)
165.66±1.965*7.50
[150.92; 180.40]
With a 95% confidence level, you'd expect that the interval [150.92; 180.40] contains the number of Asian people that would welcome a Black person in their family.
I hope it helps!