Mathematics College

At a certain college, 28% of the students major in engineering, 18% play club sports, and 8% both major in engineering and play club sports. A student is selected at random. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Given that the student is majoring in engineering, what is the probability that the student plays club sports

Answers

Answer 1

Answers: 0.286

Explanation:

Let E → major in Engineering

Let S → Play club sports

P (E) = 28% = 0.28

P (S) = 18% = 0.18

P (E ∩ S ) = 8% = 0.08

Probability of student plays club sports given majoring in engineering,

P ( S | E ) = P (E ∩ S ) ÷ P (E) = 0.08 ÷ 0.28 = 0.286

Answer 2

Final answer:

To find the probability that a student plays club sports given that they major in engineering, use conditional probability.

Explanation:

To find the probability that a student plays club sports given that they major in engineering, we need to use conditional probability.

The formula for conditional probability is:

P(A|B) = P(A ∩ B) / P(B)

In this case, A represents playing club sports and B represents majoring in engineering. We are given that P(A ∩ B) = 8% and P(B) = 28%.

Plugging these values into the formula, we get:

P(A|B) = 8% / 28% = 0.2857

So the probability that a student plays club sports given that they major in engineering is approximately 0.2857, or 28.57%.

Related Questions

The standard deviation, the range, and the interquartile range (IQR) summarize the variability of the data. a. Why is the standard deviation usually preferred over the ranges? b. Why is the IQR sometimes preferred to the standard deviation? c. What is an advantage of the standard deviation over the IQR?

Answers

Answer:

Step-by-step explanation:

a) The standard deviation is usually preferred over the range because it is calculated from all of the data and will not be impacted as much as the range when they are outliers,and the standard deviation uses all of the data.

b)The IQR sometimes referred to the standard deviation when there is an outlier because the IQR is less sensitive to this features than standard deviation.

that is the IQR is not affected by an outlier,while the standard deviation is affected by an outlier.

c)The advantage of the standard deviation over the IQR is the standard deviation takes into account the values of all observation,while the IQR uses only some of the data.

g 4. You start a dog walking business. Define to be a random variable denoting how many dogs you walkthis week. The probability mass function (pmf),f(x), o'is defined as follows:
0 1 2 3 4 5 6 7 P(X = x) 0.14 0.12 0.15 0.23 0.18 0.09 0.08 0.01
(a) Verify that f(x) is a valid probability mass function.
(b) Find the probability that you will walk at least two dogs this week.
(c) Compute the expected number of dogs you will walk this week.
(d) Compute the expected value of X2.
(e) Compute Var[x] = E(X2) – (E[X])2.

Answers

Answer:

(a) Yes, f(x) is a valid probability mass function.

(b) The probability that you will walk at least two dogs this week = 0.74.

(c) The expected number of dogs you will walk this week = 3 dogs.

(d) The expected value of X2 = 11.29

(e) Var[x] = E(X2) – (E[X])2 = 3.2811

Step-by-step explanation:

We are given with the probability mass function (pmf),f(x), o'is defined as follows:

Firstly let X = Number of dogs you walk this week

X P(X = x)

0 0.14

1 0.12

2 0.15

3 0.23

4 0.18

5 0.09

6 0.08

7 0.01

(a) Now f(x) to be a valid probability mass function, two conditions should be met :

All values should be >= 0.Sum of all probabilities must be equal to 1.

So, First condition is already met as all values are positive and for second condition = 0.14 + 0.12+ 0.15+ 0.23+ 0.18+ 0.09+ 0.08+ 0.01 = 1

Hence both the conditions are satisfied so f(x) is a valid probability mass function.

(b) Probability that we will walk at least two dogs this week = P(X>=2)

= 1 - P(X = 0) - P(X = 1) = 1 - 0.14 - 0.12 = 0.74

E(X) = [tex]\frac{\sum X\times P(X=x)}{\sum P(X=x)}[/tex] = [tex]\frac{0*0.14 + 1*0.12 + 2*0.15 + 3*0.23 + 4*0.18 + 5*0.09 + 6*0.08 + 7*0.01}{1}[/tex]

= 2.83 or 3 after rounding off.

Therefore the expected number of dogs you will walk this week are 3 dogs.

(d) The expected value of X2 [E(X2)] = [tex]\frac{\sum X^{2} \times P(X=x)}{\sum P(X=x)}[/tex]

= [tex]\frac{0^{2} *0.14 + 1^{2} *0.12 + 2^{2} *0.15 + 3^{2} *0.23 + 4^{2} *0.18 + 5^{2} *0.09 + 6^{2} *0.08 + 7^{2} *0.01}{1}[/tex] = 11.29

(e) Var[x] = E(X2) – (E[X])2

We have E(X2) = 11.29 and E(X) = 2.83

Var[x] = [tex]11.29 - (2.83)^{2}[/tex] = 3.2811.

Final answer:

The random variable X represents the number of dogs walked in a week for a dog walking business. By verifying that the sum of the probabilities equals 1, we've confirmed the pmf is valid. The probability of walking at least two dogs is 0.74, and calculations for the expected number and variance involve using the given probabilities and applying the respective formulas.

Explanation:

Understanding Probability Mass Functions

In the context of a dog walking business, let us define the random variable X as the number of dogs walked in a week. The values that X can take on are 0, 1, 2, 3, 4, 5, 6, and 7. The probabilities of these events are given by P(X = x), where x are the respective values X can take.

To verify if the given probability mass function (pmf) is valid, the sum of all probabilities P(X = x) should be equal to 1. We calculate the sum as follows:

0.14 (for X=0)0.12 (for X=1)0.15 (for X=2)0.23 (for X=3)0.18 (for X=4)0.09 (for X=5)0.08 (for X=6)0.01 (for X=7)

Adding them gives 1 which confirms it is a valid pmf.

To find the probability of walking at least two dogs this week, which is P(X ≥ 2), we need to add up the probabilities for all events where X is 2 or more:

P(X ≥ 2) = P(X=2) + P(X=3) + P(X=4) + P(X=5) + P(X=6) + P(X=7)
= 0.15 + 0.23 + 0.18 + 0.09 + 0.08 + 0.01
= 0.74

To compute the expected number of dogs walked in a week, or E[X], we use the formula:

E[X] = Σ x * P(X = x)

For E[X2], we calculate the expectation of X-squared by multiplying each value of x by itself and then by its probability, before summing these products.

Finally, the variance Var[X] of the random variable X is given by E(X2) - (E[X])2. This involves first calculating E[X2] as explained, then squaring the previously found E[X] value, and subtracting that square from E[X2].

a new toy is regularly priced at $26.99 is on sale for 3/4 off. write an expression to represent the price oc the toy,p minus 34 of the price. then, combine like terms to simplify the expression

Answers

Answer:

[tex]T(p) = \frac{1}{4}p[/tex]

The new price is $6.75

Step-by-step explanation:

The new price of the toy after the discount (T) is given by the original price (p) subtracted by the discounted amount (3/4 of p):

[tex]T(p) = p-\frac{3}{4}p \\T(p) = (1-\frac{3}{4})p\\T(p) = \frac{1}{4}p[/tex]

If the original price was $26.99, the new price is:

[tex]T= \frac{1}{4}*\$26.99\\T=\$6.75[/tex]

A sociologist is studying the effect of having children within the first two years of marriage on the divorce rate. Using hospital birth records, she selects a random sample of 200 couples that had a child within the first two years of marriage. Following up on these couples, she finds that 80 couples are divorced within five years. Use Scenario 8-4. A 90% confidence interval for the proportion p of all couples that had a child within the first two years of marriage and are divorced within five years is 0.402 ± 0.056.a. All the answers are correct.b. Based on this interval, we can clearly see that the divorce rate is well below the 50% national average for all marriages.c. At the 10% alpha level, we would reject the claim that the divorce rate is 50% for couples who had a child within the first two years of marriage.d. Based on this interval, we can clearly see that the divorce rate is between 35% and 46%.

Answers

Answer:

The correct option is (a).

Step-by-step explanation:

The hypothesis of the study can be defined as:

H₀: The divorce rate is 50% for couples who had a child within the first two years of marriage, i.e. p = 0.50

Hₐ: The divorce rate is different from 50% for couples who had a child within the first two years of marriage, i.e. p ≠ 0.50

The 90% confidence interval is: 0.402 ± 0.056 = (0.346, 0.458) ≈ (0.35, 0.46)

The confidence level is 90%, the significance level (α) is:

[tex]\alpha =1-\frac{Confidence\ level}{100}\\=1-\frac{90}{100}\\ =0.10\ or\ 10\%[/tex]

Decision Rule:

If the null hypothesis value is not contained in the 90% confidence interval then the null hypothesis will be rejected and vice-versa.

Interpretation of the Confidence interval:

The confidence interval is (35%, 46%), this implies divorce rate is less than 50% for couples who had a child within the first two years of marriage.At 10% significance level, the null hypothesis will be rejected stating that the divorce rate is different from 50% for couples who had a child within the first two years of marriage.The confidence interval clearly interprets that 90% of the divorce rate for couples who had a child within the first two years of marriage is between 35% and 46%.

Thus all the options are correct.

Final answer:

The 90% confidence interval for the proportion of couples who had a child within the first two years of marriage and are divorced within five years is 0.402 ± 0.056. Based on this interval, we can conclude that the divorce rate is between 35% and 46%.

Explanation:

Based on the given information, the sociologist selected a random sample of 200 couples who had a child within the first two years of marriage. Out of these couples, 80 were found to be divorced within five years. The 90% confidence interval for the proportion of all couples that had a child within the first two years of marriage and are divorced within five years is given as 0.402 ± 0.056.

This means that we can be 90% confident that the true proportion of couples who had a child within the first two years of marriage and are divorced within five years lies between 0.402 - 0.056 and 0.402 + 0.056.

Therefore, the correct statement based on this interval is that the divorce rate is between 35% and 46%.

Solve, graph, and give interval notation for the inequality:

4(3x − 4) < 32 AND 2x + 1 ≤ 8x + 25

Answers

Answer:

The answer to your question is

Step-by-step explanation:

Inequality 1

4(3x - 4) < 32

12x - 16 < 32

12x < 32 + 16

12x < 48

x < 48/12

x < 4

Inequality 2

2x + 1 ≤ 8x + 25

2x - 8x ≤ 25 - 1

- 6x ≤ 24

x ≥ 24/-6

x ≥ - 4

- See the graph below

- Interval [-4, 4)

Final answer:

To solve the inequality system, we divide both sides of each inequality by the respective coefficient to isolate the variable. The solutions are x < 4 and -4 ≤ x. The graph of the solution is a number line with an open circle at 4 and a shaded region to the left, and a closed circle at -4 and a shaded region to the right.

Explanation:

To solve the inequality 4(3x - 4) < 32, we divide both sides of the inequality by 4 to isolate the variable. This gives us 3x - 4 < 8. Adding 4 to both sides of the inequality gives us 3x < 12.

Finally, dividing both sides of the inequality by 3 gives us x < 4.

For the inequality 2x + 1 ≤ 8x + 25, we subtract 2x from both sides of the inequality to isolate the variable. This gives us 1 ≤ 6x + 25. Subtracting 25 from both sides of the inequality gives us -24 ≤ 6x.

Finally, dividing both sides of the inequality by 6 gives us -4 ≤ x.

The solutions to the inequality system are x < 4 and -4 ≤ x. The graph of the solution would be a number line with an open circle at 4 and a shaded region to the left, and a closed circle at -4 and a shaded region to the right.

The interval notation for the solution is (-∞, 4) and [-4, ∞).

Indicate in standard form the equation of the line passing through the given points.
L(5.0), M(0,5)

Answers

Answer:

y = - x + 5

Step-by-step explanation:

L(5.0), M(0,5)

y = mx + b

m = (5 - 0) / (0 - 5) = 5 / -5 = - 1

b = y - mx = 5 - ((-1) x 0) = 5

y = - x + 5

The perimeter of the window of the camper shell is 130 in. Find the length of one of the shorter sides of the window.
in

Answers

You can't deduce the length of a side from the perimeters of a rectangle.

Say that [tex]s[/tex] and [tex]S[/tex] are, respectively, the short and long side of the rectangle.

So, we know that

[tex]2s+2S=130 \iff 2(s+S)=130 \iff s+S=65[/tex]

But we can't solve exactly for [tex]s[/tex] nor for [tex]S[/tex], unless more information is given.

Final answer:

The length of one of the shorter sides of the window cannot be explicitly determined without additional information. However, considering the window has a rectangular shape, the length of the shorter side should be less than half of the total perimeter, in this case, less than 65 inches.

Explanation:

To find the length of one of the shorter sides of the window, we must first understand that the perimeter of a rectangle is calculated by the formula: 2*length + 2*width = Perimeter.

However, the problem doesn't specify the dimensions of the box, but it does tell us that one pair of sides (the length and the width) are not equal. This suggests that the window is a rectangle. If we assume that the length of the window is longer than the width (length > width), then the shorter sides of the rectangle (the widths) will be of equal length.

Without more details, we can't calculate the exact measurement of the length of one of the shorter sides, but we can say that it should be less than half of the total perimeter, which is less than 65 inches.

Learn more about Perimeter of a Rectangle here:

https://brainly.com/question/29595517

#SPJ12

A 25-foot ladder is leaning against a house with the base of the ladder 5 feet from the house. How high up the house does the ladder reach? Round to the nearest tenth of a foot. The ladder reaches feet up the side of the house

Answers

Answer: 24 feet

Step-by-step explanation:

By using Pythagoras rule:

Let x be the high up the house does the ladder reached.

X^2 + 5^2= 25^2

X^2 = 25^2 - 5^2

x^2 = 625 - 25

x^2 = 600

Square both side

x = sqrt(600)

x= 24.495

x = 24 feet

In one week, Mohammed can knit 5 sweaters or bake 240 cookies. In one week Aisha can knit 15 sweaters or bake 480 cookies. Mohammed's opportunity cost knitting one sweater is: A. 480 cookies. B. 240 cookies. C. 48 sweaters. D. 1/48 of a cookie E. 48 cookies.

Answers

Answer:

We conclude, Mohammed's opportunity cost knitting one sweater is 48 cookies.

Step-by-step explanation:

We have that Mohammed can knit 5 sweaters or bake 240 cookies.

In one week Aisha can knit 15 sweaters or bake 480 cookies.

We calculate how much is Mohammed's opportunity cost knitting one sweater. We get

\frac{240}{5}= 48.

We conclude, Mohammed's opportunity cost knitting one sweater is 48 cookies.

Two students, X and Y, forgot to put their names on their exam papers. The professor knows that these two students do well on the exam with probabilities 0.8 and 0.4, respectively. After grading, the professor notices that X and Y forgot to put their names on their exams. One of their exams was done well and the other was done poorly. Given this information, and assuming that students worked independently of each other, what is the probability that the good exam belongs to student X

Answers

Answer:

The probability that the good exam belongs to student X is 0.8571.

Step-by-step explanation:

It is provided that the probability that X did well in the exam is, P (X) = 0.90 and the probability that X did well in the exam is, P (Y) = 0.40,

Compute the probability that exactly one student does well in the exam as follows:

[tex]P(Either\ X\ or\ Y\ did\ well)=P(X\cap Y^{c})+P(X^{c}\cap Y)\\=P(X)P(Y^{c})+P(X^{c})P(Y)\\=P(X)[1-P(Y)]+[1-P(X)]P(Y)\\=(0.80\times0.60)+(0.20\times0.40)\\=0.56[/tex]

Then the probability that X is the one who did well in the exam is:

[tex]P(X\ did\ well\ in\ the\ exam)=\frac{P(X\cap Y^{c})}{P(X\cap Y^{c})+P(X^{c}\cap Y)}\\ =\frac{P(X)[1-P(Y)]}{P(X\cap Y^{c})+P(X^{c}\cap Y)} \\=\frac{0.80\times0.60}{0.56}\\=0.857143\\\approx0.8571[/tex]

Thus, the probability that the good exam belongs to student X is 0.8571.

In a study of environmental lead exposure and IQ, the data was collected from 148 children in Boston, Massachusetts. Their IQ scores at age of 10 approximately follow a normal distribution with mean of 115.9 and standard deviation of 14.2. Suppose one child had an IQ of 74. The researchers would like to know whether an IQ of 74 is an outlier or not.

Calculate the lower fence for the IQ data, which is the lower limit value that the IQ score can be without being considered an outlier. Keep a precision level of two decimal places for the lower fence.

Answers

Answer:

a) Lower inner fence = 77.6168 = 77.62 to 2 d.p

Lower outer fence = 48.9044 = 48.90 to 2 d.p

b) The probability of obtaining an IQ score value of 74 or less is P(x ≤ 74) is 0.00159

Step-by-step explanation:

Lower inner and outer fences are used to illustrate or write off extreme values of a data set (the outliers).

Lower inner fence = Q₁ – (1.5 × IQR)

Lower outer fence = Q₁ – (3 × IQR)

Q₁ = 25th percentile = lower quartile

IQR = Inter quartile Range = Q₃ - Q₁

Q₃ = 75th percentile = upper quartile

To calculate Q₁ for a normal distribution with only mean and standard deviation known,

We need the standardized score whose probability is 0.25 P(z) = 0.25

From the normal distribution table

z = (± 0.674)

z = (x - xbar)/σ

x = the value in the data we're interested in,

xbar = mean = 115.9

σ = standard deviation = 14.2

Lower quartile corresponds to (z = - 0.674)

- 0.674 = (x - 115.9)/14.2

Q₁ = X = 106.3292

The upper quartile, Q₃ corresponds to z = (+0.674)

Q₃ = 125.4708

IQR = 125.4708 - 106.3292 = 19.1416

Lower inner fence = Q₁ – (1.5 × IQR)

Lower outer fence = Q₁ – (3 × IQR)

Lower inner fence = 106.3292 - (1.5 × 19.1416) = 106.3292 - 28.7124 = 77.6168

Lower outer fence = 106.3292 – (3 × 19.1416) = 48.9044

b) The probability of obtaining an IQ score value of 74 or less is P(x ≤ 74)

We standardize 74 by obtaining its z-score

z = (x - xbar)/σ

z = (74 - 115.9)/14.2 = - 2.95

P(x ≤ 74) = P(z ≤ -2.95) = 0.00159 (Obtained from normal distribution tables)

Final answer:

The lower fence for the IQ data, which determines whether an IQ score is an outlier, is calculated as the mean minus two times the standard deviation. In this case, the lower fence is 87.5, which makes an IQ score of 74 an outlier as it falls significantly below this threshold.

Explanation:

To determine if an IQ score is an outlier, we often use the interquartile range (IQR) and calculate the fences. However, since the data is approximately normally distributed and we have the mean and standard deviation, we can also consider an IQ score to be an outlier if it falls more than two standard deviations from the mean. In this question, we do not have the IQR, so we'll use standard deviations to calculate the outlier threshold.

The mean IQ score is 115.9 and the standard deviation is 14.2. An outlier is typically defined as a value that is more than two or three standard deviations away from the mean. These thresholds are sometimes called the outer fences in statistical outlier detection. Using two standard deviations, we can calculate the lower limit as follows:

Lower Limit = Mean - 2 × Standard Deviation
Lower Limit = 115.9 - 2(14.2)
Lower Limit = 115.9 - 28.4
Lower Limit = 87.5

Therefore, an IQ score of 74 is considerably lower than the lower limit of 87.5, suggesting that it could indeed be considered an outlier.

As a result of discharges from local dry cleaner, dinitrotoluene concentration in the groundwater is 8 mg/L. RfD for dinitrotoluene is 2.0 x 10-3 mg/kg-day. The average 70 Kg person drinks 2L/day water. The hazard ratio is most nearly:

Answers

Answer:

114.3

Step-by-step explanation:

If a 70kg person ingests 2L of water per day containing 8 mg/L of dinitrotoluene, the concentration of dinitrotoluene on that person's body is:

[tex]C=2\frac{L}{day}*8\frac{mg}{L} *\frac{1}{70\ kg}\\C=0.22857\frac{mg}{kg-day}[/tex]

The hazard ratio is defined by dividing the intake dosage (C) by the reference dose (RfD)

[tex]H=\frac{0.22857}{2*10^{-3}}\\H=114.3[/tex]

The hazard ratio is most nearly 114.3.

A woman sued a computer keyboard manufacturer, charging that her repetitive stress injuries were caused by the keyboard. The injury awarded about $3.5 million for pain and suffering, but the court then set aside that award as being unreasonable compensation. In making this determination, the court identified a "normative" group of 27 similar cases and specified a reasonable award as one within two standard deviations of the mean of the awards in the 27 cases. The 27 awards were (in $1000s) 36, 62, 73, 114, 139, 140, 148, 154, 238, 290, 340, 410, 600, 750, 750, 750,1050, 1100, 1135, 1150, 1200, 1200, 1250, 1578, 1700, 1825, and 2000, from which?xi = 20,182, ?xi2 = 24,656,384.What is the maximum possible amount that could be awarded under the two-standard-deviation rule? (Round your answer to the nearest whole number.)

Answers

Answer:

variance = (27*24656384-20182^2)/(27*26) =368104.3

standard devaition SD= sqrt(368104.3) =606.716

maximum possible amount that could be awarded under the two-standard-deviation rule = mean +2*SD

= (20182/27)+(2*606.716)

= 1960.913

=$1960913

Calculate the infant mortality rate (per 1,000 live births) from the following data:
a. Number of infant deaths under 1 year in the United States during 1991 = 36,766
b. Number of live births during 1991 = 4,111,000

Answers

Answer:

ahsbbsnssisiskak sjsnsnaan

A sports statistician is interested in determining if there is a relationship between the number of home team and visiting team losses and different sports. A random sample of 526 games is selected and the results are given below. Find the critical value chi Subscript alpha Superscript 2 to test the claim that the number of home team and visiting team losses is independent of the sport. Use alphaequals0.01. Round to three decimal places.

Answers

Answer:

The critical value would be: [tex]\chi^2_{crit}=11.345[/tex] and we use the following excel code to find it: "=CHISQ.INV(1-0.01,3)"

[tex]p_v = P(\chi^2_{3} >3.29)=0.349[/tex]

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(3.29,3,TRUE)"

Since the p value is higher than the significance level we FAIL to reject the null hypothesis at 5% of significance that the two variables are independent.

Step-by-step explanation:

Previous concepts

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

Assume the following dataset:

F B S Bs Total

home wins 39 156 25 83 303

Visitor wins 31 98 19 75 223

Total 70 254 44 158 526

We need to conduct a chi square test in order to check the following hypothesis:

H0: The number of home team and visiting team losses is independent of the sport.

H1: The number of home team and visiting team losses is dependent of the sport.

The level of significance assumed for this case is [tex]\alpha=0.01[/tex]

The statistic to check the hypothesis is given by:

[tex]\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}[/tex]

The table given represent the observed values, we just need to calculate the expected values with the following formula [tex]E_i = \frac{total col * total row}{grand total}[/tex]

And the calculations are given by:

[tex]E_{1} =\frac{70*303}{526}=40.32[/tex]

[tex]E_{2} =\frac{254*303}{526}=146.32[/tex]

[tex]E_{3} =\frac{44*303}{526}=25.35[/tex]

[tex]E_{4} =\frac{158*303}{526}=91.02[/tex]

[tex]E_{5} =\frac{70*223}{526}=29.68[/tex]

[tex]E_{6} =\frac{254*223}{526}=107.68[/tex]

[tex]E_{7} =\frac{44*223}{526}=18.65[/tex]

[tex]E_{8} =\frac{158*223}{526}=66.98[/tex]

And the expected values are given by:

F B S Bs Total

home wins 40.32 146.32 25.35 91.02 303

Visitor wins 29.68 107.68 18.65 66.98 223

Total 70 254 44 158 526

And now we can calculate the statistic:

[tex]\chi^2 = \frac{(39-40.32)^2}{40.32}+\frac{(156-146.32)^2}{146.32}+\frac{(25-25.35)^2}{25.35}+\frac{(83-91)^2}{91}+\frac{(31-29.68)^2}{29.68}+\frac{(98-107.68)^2}{107.68}+\frac{(19-18.65)^2}{18.65}+\frac{(75-66.98)^2}{66.98} =3.29[/tex]Now we can calculate the degrees of freedom for the statistic given by:

[tex]df=(rows-1)(cols-1)=(4-1)(2-1)=3[/tex]

The critical value would be: [tex]chi^2_{crit}=11.345[/tex] and we use the following excel code to find it: "=CHISQ.INV(1-0.01,3)"

And we can calculate the p value given by:

[tex]p_v = P(\chi^2_{3} >3.29)=0.349[/tex]

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(3.29,3,TRUE)"

Since the p value is higher than the significance level we FAIL to reject the null hypothesis at 5% of significance that the two variables are independent.

In an experimental study on friendliness and tipping, every alternate customer to whom the waiter is extra friendly toward are referred to as the...A. Control group
B. Experimental group
C. Nonexperimental group
D. Dependent group

Answers

Answer:

A. Control group

Step-by-step explanation:

A control group is a group in an experiment or study that does not receive experimental procedure during such that it is then used as a benchmark to measure how the other tested subjects do.

An experimental group is a group in an experiment or study that receives an experimental procedure. The values of gotten from the test are recorded and the effect of independent variables on the dependent variables are determined.

Control experiment can be used to determine whether or not the customers are friendly. Experimental group will be the customers whom the waiter is extra friendly toward. The control group will be the alternate customer whom the waiter is not extra friendly toward.

Find the direction cosines and direction angles of the given vector. (Round the direction angles to two decimal places.) a = 5, 9, 3 cos(α) = cos(β) = cos(γ) = α = ° β = ° γ = °

Answers

Answer:

Step-by-step explanation:

given is a vector as (5,9,3)

a = (5,9,3)

To find out direction cosines

First let us calculate modulus of vector a

[tex]||a|| =\sqrt{5^2+9^2+3^2} \\=\sqrt{25+81+9} \\=\sqrt{115}[/tex]

Direction ratios are (5,9,3)

Magnitude of vector a = [tex]\sqrt{115}[/tex]

So direction cosines would be

[tex](\frac{5}{\sqrt{115} } ,\frac{9}{\sqrt{115} },\frac{3}{\sqrt{115} })[/tex]

Angles would be

[tex](\alpha, \beta, \gamma) = arccos ((\frac{5}{\sqrt{115} } ,\frac{9}{\sqrt{115} },\frac{3}{\sqrt{115} })[/tex]

=cos inverse (0.4662, 0.8393, 0.2798)

= (62.21, 32.93,32,94)

Suppose that the times required for a cable company to fix cable problems in the homes of its customers are uniformly distributed between 40 minutes and 65 minutes. What is the probability that a randomly selected cable repair visit falls within 2 standard deviations of the mean?

Answers

Answer: 1

Step-by-step explanation:

If a random variable x is uniformly distributed in [a,b] the

Mean = [tex]\dfrac{a+b}{2}[/tex]

Standard deviation : [tex]\sqrt{\dfrac{(b-a)^2}{12}}[/tex]

Let x = Times required for a cable company to fix cable problems

As per given.

x is uniformly distributed between 40 minutes and 65 minutes.

Then , mean = [tex]\dfrac{65+40}{2}=52.5[/tex] minutes

Standard deviation : [tex]\sqrt{\dfrac{(65-40)^2}{12}}\approx7.22[/tex]minutes

Consider , P (mean- 2(Standard deviation) < X < mean+2(Standard deviation) )

= P(52.5-2(7.22)< X < 52.5+2(7.22))

=P(38.06 <X < 66.94 ).

But x lies between 40 minutes and 65 minutes.

Also, [40 minutes, 65 minutes]⊂ [38.06 minutes , 66.94 minutes]

Therefore ,P(38.06 <X < 66.94 ) =1

∴ The probability that a randomly selected cable repair visit falls within 2 standard deviations of the mean is 1.

The empirical rule states that 95% of the distribution is within 2 standard deviations of the mean,

Z score

Z score is used to determine by how many standard deviations the raw score is above or below the mean. The z score is given by:

z = (x - μ)/σ

where x is the raw score, μ is the mean and σ is the standard deviation.

The empirical rule states that 95% of the distribution is within 2 standard deviations of the mean.

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An experiment was performed upon rats to investigate the effect of ingesting Alar (a chemical sprayed on apple trees to keep fruit from dropping before ripe) upon subsequent cancer rates.
The following variables were measured:
gender (0=female, 1=male); weight (g); dose of Alar (nil, low, high); and number of tumors.
Which of the following is FALSE?

A) Gender is categorical; dose is ordinal
B) Gender is discrete; weight is continuous
C) Number of tumors is categorical
D) Dose is discrete
E) Weight is continuous

Answers

Answer:

Option C and D are false

Step-by-step explanation:

All the mentioned option are correct in the given scenario except option C and D.

The reason is that dose is categorized as nil, low and high so, dose is categorical variable. Also, number of tumors is quantitative variable because it can be meaningfully interpreted in numerical form. The number if tumors is discrete quantitative variable.

Now consider all options

A) Gender is categorical ; dose is ordinal

This option is true because gender can be categorized into male and female and also dose is ordinal because it has order i.e. nil,low and high.

B) Gender is discrete; weight is continuous

This option is false because gender can be a discrete variable and weight is continuous variable because it is measurable. So, the statement is true.

Option C and D are already discussed an option E is discussed in option B.

Final answer:

The false statement is C) Number of tumors is categorical. The number of tumors is a discrete variable that involves countable values, not a categorical variable.

Explanation:

In this experiment, we have different types of variables. The statement C) Number of tumors is categorical is false. The number of tumors is actually a discrete variable as it involves countable values. The other statements are correct: A) Gender is categorical; dose is ordinal, as gender is divided into male and female, which is a categorical classification, and the dose is ranked as nil, low, high which makes it an ordinal variable. Statement B) Gender is discrete; weight is continuous is also correct because gender is a discrete variable (only two possible values, male or female), and weight is a continuous variable as it can take any value within a certain range. Statement D) Dose is discrete is correct, as the dose can only take certain values (nil, low, high) it is considered a discrete variable. Lastly, E) Weight is continuous is accurate as Weight can take any value within a range and it involves measurement.

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Exercise 1.28. We have an urn with m green balls and n yellow balls. Two balls are drawn at random. What is the probability that the two balls have the same color? (a) Assume that the balls are sampled without replacement. (b) Assume that the balls are sampled with replacement. (c) When is the answer to part (b) larger than the answer to part (a)? Justify your answer. Can you give an intuitive explanation for what the calculation tells you?

Answers

Answer:

Step-by-step explanation:

given that we have an urn with m green balls and n yellow balls. Two balls are drawn at random.

a) Assume that the balls are sampled without replacement.

m green and n yellow balls

For 2 balls to be drawn at the same colour

no of ways = either 2 green or 2 blue = mC2+nC2

Total no of ways = (m+n)C2

Prob =

= [tex]\frac{mC2 +nC2}{(m+n)C2} \\=\frac{m(m-1)+n(n-1)}{(m+n)(m+n-1)}[/tex]

=[tex]\frac{m^2+n^2-m-n}{(m+n)(m+n-1)}[/tex]

B) Assume that the balls are sampled with replacement

In this case, probability for any draw for yellow or green will be constant as

n/M+n or m/m+n respectively

Reqd prob = [tex](\frac{m}{m+n} )^2 +(\frac{n}{m+n} )^2[/tex]

=[tex]\frac{m^2+n^2}{(m+n)^2}[/tex]

c) Part B prob will be more than part a because with replacement prob is more than without replacement.

II time drawing same colour changes to m-1/.(m+n-1) if with replacement but same as m/(m+n) without replacement

[tex]\frac{m}{m+n} >\frac{m-1}{m+n-1} \\m^2+mn-m>m^2+mn-m-n\\n>0[/tex]

Since n>0 is true always, b is greater than a.

Final answer:

The question explores the concept of probability within scenarios of drawing balls of different colors from an urn, with and without replacement. It explores how the number of balls left in the urn changes the likelihood of drawing two balls of the same color. The answer is calculated using mathematical odds and conditions, showing that replacement affects probability especially when the total number of items (balls in this case) is small.

Explanation:

The subject of this question is probability, specifically conditional probability and probability with and without replacement. Here are the calculations needed to answer the question:

(a) When the balls are drawn without replacement, the probability that the two balls drawn have the same color is the sum of the probability of drawing two green balls and the probability of drawing two yellow balls. The probability of drawing two green balls is (m/(m+n)) * ((m-1)/(m+n-1)). Similarly, the probability of drawing two yellow balls is (n/(m+n)) * ((n-1)/(m+n-1)). The sum of these two probabilities gives the required probability. (b) When the balls are drawn with replacement, the same logic applies; however, since the balls are replaced, the denominator term doesn't decrease for the second draw. Thus, the probability of drawing two green balls is (m/(m+n)) * (m/(m+n)), and the probability of two yellow balls is (n/(m+n)) * (n/(m+n)). (c) The answer to part (b) becomes larger than the answer to part (a) when m and n are small numbers. This is because, when m and n are small, the probability of drawing a similarly colored ball in the second draw becomes more significant if the ball is replaced after the first draw, compared to if it is not replaced, causing the probability with replacement to be higher.

In a nutshell, the calculation for probability tells us how likely an outcome is, given the mathematical odds and conditions (in this case, if the balls are replaced or not after drawing).

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Find the distance between the points (-5, -10) and (2, 4).

Math item stem image
CLEAR CHECK

4.58

12.12

15.65

21

Answers

Answer:

15.65

Step-by-step explanation:

Suppose we have two points:

[tex]A = (x_{1}, y_{1})[/tex]

[tex]B = (x_{2}, y_{2})[/tex]

The distance between these points is:

[tex]D = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}[/tex]

So, for points (-5, -10) and (2, 4)

[tex]D = \sqrt{(2 - (-5))^{2} + (4 - (-10))^{2}}[/tex]

[tex]D = \sqrt{7^{2} + 14^{2}[/tex]

[tex]D = 15.65[/tex]

So the correct answer is:

15.65

A dead body was found within a closed room of a house where the temperature was a constant 65° F. At the time of discovery the core temperature of the body was determined to be 85° F. One hour later a second measurement showed that the core temperature of the body was 80° F. Assume that the time of death corresponds to t = 0 and that the core temperature at that time was 98.6° F. Determine how many hours elapsed before the body was found.

Answers

Answer:

1 hr 52 minutes

Step-by-step explanation:

As per Newton law of cooling we have

[tex]T(t) = T_s +(T_0-T_s)e^{-kt}[/tex]

where T0 is the initial temperature of the body

Ts = temperature of surrounding

t = time lapsed

k = constant

Using this we find that T0 = 98.6 : Ts= 65

Let x hours be lapsed before the body was found.

Then we have

[tex]T(x) = 65 +(98.6-65)e^{-kx} = 85\\e^{-kx}=\frac{20}{33.8} =0.5917[/tex]

Next after 1 hour temperature was 80

[tex]T(x+1) = 65+33.6(e^{-k(x+1)}=80\\e^{-k(x+1) =0.4464[/tex]

Dividing we get

[tex]e^k = 1.325408\\k = 0.2817[/tex]

Substitute this in

[tex]e^{-kx} =0.5917\\x=\frac{ln 0.5917}{-k} \\=1.863[/tex]

approximately 1 hour 52 minutes have lapsed.

According to a recent report, 60% of U.S. college graduates cannot find a full time job in their chosen profession. Assume 57% of the college graduates who cannot find a job are female and that 18% of the college graduates who can find a job are female. Given a male college graduate, find the probability he can find a full time job in his chosen profession? (See exercise 58 on page 220 of your textbook for a similar problem.)

Answers

Answer:

There is a 55.97% that a male can find a full time job in his chosen profession.

Step-by-step explanation:

We have these following probabilities:

A 60% probability that a college graduates cannot find a full time job in their chosen profession.

A 40% probability that a college graduates can find a full time job in their chosen profession.

57% of the college graduates who cannot find a job are female

43% of the college graduates who cannot find a job are male

18% of the college graduates who can find a job are female

82% of the college who can find a job are male.

Given a male college graduate, find the probability he can find a full time job in his chosen profession?

The total males are 43% of 60%(Those who cannot find a job) and 82% of 40%(Those who can find a job). So the percentage of males is [tex]P(M) = 0.43*0.60 + 0.82*0.40 = 0.586[/tex]

Those who are males and find a job in their chosen profession are 82% of 40%. So [tex]P(M \cap J) = 0.82*0.40 = 0.328[/tex]

[tex]P = \frac{P(M \cap J)}{P(M)} = \frac{0.328}{0.586} = 0.5597[/tex]

There is a 55.97% that a male can find a full time job in his chosen profession.

Suppose that the probability of a baseball player getting a hit in an at-bat is 0.3089. If the player has 25 at-bats during a week, what's the probability that he gets greater than 9 hits?

Answers

Answer:

[tex] P(X>9) = 0.3593[/tex]

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest, on this case we now that:

[tex]X \sim Binom(n=25, p=0.3089)[/tex]

The probability mass function for the Binomial distribution is given as:

[tex]P(X)=(nCx)(p)^x (1-p)^{n-x}[/tex]

For this case we want this probability:

[tex] P(X >9)[/tex]

And we can use the complement rule like this:

[tex] P(X>9) = 1-P(X \leq 8)= 1-[P(X=0) + P(X=1) +....+P(X=8)] [/tex]And we can find the individual probabilities like this:

[tex] P(X=0) =(25C0)(0.3089)^0 (1-0.3089)^{25-0} =0.0000974[/tex]

[tex] P(X=1) =(25C1)(0.3089)^1 (1-0.3089)^{25-1} =0.0011[/tex]

[tex] P(X=2) =(25C2)(0.3089)^2 (1-0.3089)^{25-2}=0.00584[/tex]

[tex] P(X=3) =(25C3)(0.3089)^3 (1-0.3089)^{25-3}= 0.02[/tex]

[tex] P(X=4) =(25C4)(0.3089)^4 (1-0.3089)^{25-4}=0.049[/tex]

[tex] P(X=5) =(25C5)(0.3089)^5 (1-0.3089)^{25-5}=0.092[/tex]

[tex]P(X=6)=(25C6) (0.3089)^6 (1-0.3089)^{25-6} = 0.138[/tex]

[tex] P(X=7) =(25C7)(0.3089)^7 (1-0.3089)^{25-7}=0.167[/tex]

[tex] P(X=8) =(25C8)(0.3089)^8 (1-0.3089)^{25-8}=0.168[/tex]

And in order to do the operations we can use the following excel code:

"=1-BINOM.DIST(8,25,0.3089,TRUE)"

And we got:

[tex] P(X>9) = 0.3593[/tex]

Suppose an arrow is shot upward on the moon with a velocity of 44 m/s, then its height in meters after tt seconds is given by h(t)=44t−0.83t2h(t)=44t-0.83t2. Find the average velocity over the given time intervals.

a. [3, 4]:
b. [3, 3.5]:
c. [3, 3.1]:
d. [3, 3.01]:
e. [3, 3.001]

Answers

Answer:

a. 38.19m/s

b. 38.605m/s

c. 38.937m/s

d. 39.0117m/s

e. 39.01917m/s

Step-by-step explanation:

The average velocity is defined as the relationship between the displacement that a body made and the total time it took to perform it. Mathematically is given by the next formula:

[tex]v_a_v_g = \frac{\Delta x}{\Delta t} =\frac{x_f-x_i}{t_f-t_i}[/tex]

Where:

[tex]x_f=Final\hspace{3}distance\hspace{3}traveled\\x_i=Initial\hspace{3}distance\hspace{3}traveled\\t_f=Final\hspace{3}time\hspace{3}interval\\t_i=Initial\hspace{3}time\hspace{3}interval[/tex]

a. Let's find h(3) and h(4) using the data provided by the problem:

[tex]h(3)=44(3)-0.83(3^2)=124.53=x_i\\h(4)=44(4)-0.83(4^2)=162.72=x_f[/tex]

The average velocity over the interval [3, 4] is :

[tex]v_a_v_g=\frac{162.72-124.53}{4-3} =38.19m/s[/tex]

b. Let's find h(3.5) using the data provided by the problem:

[tex]h(3.5)=44(3.5)-0.83(3.5^2)=143.8325=x_f[/tex]

The average velocity over the interval [3, 3.5] is :

[tex]v_a_v_g=\frac{143.8325-124.53}{3.5-3} =38.605m/s[/tex]

c. Let's find h(3.1) using the data provided by the problem:

[tex]h(3.1)=44(3.1)-0.83(3.1^2)=128.4237=x_f[/tex]

The average velocity over the interval [3, 3.1] is :

[tex]v_a_v_g=\frac{128.4237-124.53}{3.1-3} =38.937m/s[/tex]

d. Let's find h(3.01) using the data provided by the problem:

[tex]h(3.1)=44(3.01)-0.83(3.01^2)=124.920117=x_f[/tex]

The average velocity over the interval [3, 3.01] is :

[tex]v_a_v_g=\frac{124.920117-124.53}{3.01-3} =39.0117m/s[/tex]

e. Let's find h(3.001) using the data provided by the problem:

[tex]h(3.001)=44(3.001)-0.83(3.001^2)=124.5690192=x_f[/tex]

[tex]v_a_v_g=\frac{124.5690192-124.53}{3.001-3} =39.01917m/s[/tex]

The average velocity over a given time [3,4] is 38.19 m/sec, the average velocity over a given time [3,3.5] is 38.605 m/sec, the average velocity over a given time [3,3.1] is 38.937 m/sec and this can be determined by using the given data.

Given :

Suppose an arrow is shot upward on the moon with a velocity of 44 m/s, then its height in meters after tt seconds is given by [tex]\rm h(t) = 44t-0.83t^2[/tex].

a) [3,4]

At time t = 3 and 4, the value of h(t) is given below: