You buy three tickets to play the "Mega Big Time Jackpot" in which the prize is $200,000. The chance any ticket wins is 1 out of 614,679 and is independent of any other ticket winning, what is the chance you lose all three times? Round to two decimal places.


a. 0.95
b. 1.00
c. 0.99d. 0.00

Answer:

Mean = 5; Standard Deviation: 1.5811

Step-by-step explanation:

Given Data:

number of times coin flipped = n = 10;

probability of each side of coin = p = 0.5;

Here mean is the product of number of times coin flipped and probability of each

m = n*p =10*0.5 = 5

Standard deviation is obtained by taking square root of product of n,p,q

St. Dev= [tex]\sqrt{npq}[/tex] = [tex]\sqrt{10*0.5*0.5}[/tex] = 1.5811

We have:

           Mean = 5         ;          Standard Deviation = 1.5811

Answer: D. Excess Kurtosis

Step-by-step explanation:

A fat tailed distribution is a kind of probability distribution that exhibits excess kurtosis because it means the resulting numbers from the probability distribution are on a large scale power increment or very small/ slow decreeing order. This makes the graph on the distribution literally fat tailed and makes skewness in such distribution data extremely difficult to ascertain.

Fat tails in a distribution signify excess kurtosis, which signifies more extreme values or outliers than in a normal distribution. Neither positive skewness, negative skewness, nor a kurtosis of zero signify a distribution's 'fat tails'.

If a distribution has 'fat tails', it represents 'excess kurtosis'. This term is used to describe a distribution of data that features tails that are fatter and longer than in a normal distribution. This often means the distribution exhibits more extreme values or outliers. When a distribution has excess kurtosis, it has strong outliers.

Positive skewness, negative skewness, and a kurtosis of zero have no correlation with 'fat tails'. While skewness refers to the asymmetry of a distribution, and a kurtosis of zero refers to a normal distribution, neither of these refer to the concept of 'fat tails'.

So, Fat tails in a distribution signify excess kurtosis, not positive skewness, negative skewness, or a kurtosis of zero.

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Answer:

a) decreases at interval (-∞,5) and  increases at (5,∞)

b)  is convave down at interval (-∞,6) an up at interval (6,∞)

c) f(x) has an inflexion point at x=6

Step-by-step explanation:

a) for the function

f(x) = (4 − x)*e^(−x)

then the derivative of f(x) indicates if the function decreases or increases. Thus

f'(x) =df(x)/dx = -e^(−x) -(4 − x)*e^(−x)= (x-5)*e^(−x)

since e^(−x) is always positive , then

f'(x) < 0 for x<5 → f(x) decreases when x<5 ( interval (-∞,5) )

f'(x) > 0 for x>5 → f(x) increases when x>5 ( interval (5,∞) )

f'(x) = 0 for x=5 → f(x) has a local minimum ( since first decreases and then increases)

b) the concavity is found with the second derivative of f(x) , then

f''(x) =d²f(x)/(dx)² = e^(−x) - (x-5)*e^(−x) = (6-x)*e^(−x)

then

f''(x) < 0 for x>6 → f(x) is convave up for x>6 ( interval (6,∞) )

f'(x) > 0 for x<6 → f(x) is concave down  when x<6 ( interval (-∞,6) )

f'(x) = 0 for x=6 → f(x) has an inflection point at x=6

The function f(x) = (4 - x)e^-x is decreasing on the interval (-∞, 1), increasing on (1, ∞), concave up on (2, ∞), concave down on (-∞, 2), and has its point of inflection at [tex](2, 2e^-^2).[/tex]

To find the intervals of increase or decrease, we first need to find the derivative f'(x) of the function f(x) = (4 - x)e-x. Using the product rule and chain rule, we find [tex]f'(x) = e-x - (x - 4)e-x.[/tex]

Setting f'(x) to zero and solving for x, we find the critical points x = 1 and x = 4. Using test points, we determine that the function is decreasing on (-∞, 1) and increasing on (1, ∞).

Next, we find the second derivative [tex]f''(x) = -2e-x + 2(x - 4)e-x[/tex] to determine concavity. Setting f''(x) equal to zero and solving for x, we find x = 2. Using test points, we find that the function is concave up on (2, ∞) and concave down on (-∞, 2). Since the function changes concavity at x = 2, this is the point of inflection.

Substituting x = 2 into the original function f(x), we find y = 2e-2, so the point of inflection is (2, 2e-2).

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Answer:

[tex]16c+15[/tex]

Step-by-step explanation:

we have the expression

[tex]10c+6c+15[/tex]

step 1

We can simplify the expression by combining like terms. That is, the terms with the same variable

[tex](10c+6c)+15[/tex]

[tex]16c+15[/tex]

Answer: 16c+15

Step-by-step explanation:

Step 1

Write down the given expression (10c+6c)+15

Step 2

10+6 = 16c

So, 16c+15

Hope this helps! (✿◡‿◡)

Answer:

The passengers have an average of 8.15 minutes to wait in line

Step-by-step explanation:

Using Little's law

Average Inventory = Average Flow Time * Average Flow Rate

Average Inventory = 220 passengers

Average Flow Rate = 27

Average Flow time =?

So,

220 = Average Flow Time * 27

Average Flow Time = 220/27

Average Flow Time = 8.14814814814

Average Flow Time = 8.15 --------- Approximated

So the average wait time for a passenger is 8.15 minutes

The average passenger had to wait in line for approximately 0.123 minutes.

To find the average waiting time, we need to divide the total waiting time by the number of passengers. The total waiting time can be calculated by multiplying the average number of passengers waiting (27) by the time period (1 hour). The average waiting time is then found by dividing this total waiting time by the number of passengers checked in (220).

Average waiting time = (Average number of passengers waiting × Time period) / Number of passengers checked in

Therefore, the average passenger had to wait in line for approximately 0.123 minutes.

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Final answer:

The correct answer is c. The three groups do not add up to 100%. A pie chart is used to display the parts of a whole, where each category represents a proportion of the total. In this case, the categories of cigarette use, smokeless tobacco use, illicit drug use, and pain reliever/sedative use do not add up to 100% when combined. As a result, a pie chart would not accurately represent the data.

Explanation:

The correct answer is c. The three groups do not add up to 100%.

A pie chart is used to display the parts of a whole, where each category represents a proportion of the total. In this case, the categories of cigarette use, smokeless tobacco use, illicit drug use, and pain reliever/sedative use do not add up to 100% when combined. As a result, a pie chart would not accurately represent the data.

Answer:

Step-by-step explanation:

Given that special deck of cards has 20 cards. Nine are green, seven are blue, and four are red. When a card is picked, the color of it is recorded. An experiment consists of first picking a card and then tossing a coin

A) Sample space will have Green, Head,  or Green, Tail .... Red, head, red, tail

No of elements in sample space = no of colours x no of outcomes in coin toss

= 4x2 = 8

B) A= getting (RT)

P(A) = Prob of getting red card and tail on coin

= P (R) *P(T)

=[tex]\frac{4}{20} *\frac{1}{2} \\=\frac{1}{10}[/tex]

C) B be the event that a green or blue is picked, followed by landing a tail on the coin toss

B = getting green card and tail

Getting green card tail is mutally exclusive with red card and tail as there is no common element between green and blue.

D) C= red or green card is picked followed by tail.

Here A and C have a common element as getting red and tail.  So not mutually exclusive

Final answer:

The sample space of the card and coin toss experiment consists of 40 elements. The probability of picking a red card followed by a tail coin toss is 0.1. Events A and B, as well as events A and C, are not mutually exclusive.

Explanation:

Probability in a Card and Coin Toss Experiment

When dealing with a special deck of cards and a subsequent coin toss, multiple steps are involved in determining outcomes and probabilities. As for the given student's question:

A). The sample space contains multiple elements based on the card colors and the side of the coin: each of the 20 cards can result in either heads or tails, creating a total of 40 possible outcomes.

B). For event A, where a red card is picked followed by a tails on the coin toss, the probability (P(A)) is calculated by dividing the number of successful outcomes by the number of total possibilities, resulting in P(A) = 4 (red cards) * 1 (tails outcome) / 40 (total outcomes) = 0.1.

C). Event A and event B are not mutually exclusive because event B involves picking either a green or blue card and also getting tails, which does not overlap with the specifics of event A.

D). Similarly, events A and C are not mutually exclusive. Although both involve picking a red card and landing a tail, event C also includes picking a green card, which does not interfere with the occurrence of event A.

The zeroth derivative of a function [tex]y(x)[/tex] is simply the function itself, so the zeroth order linear ODE takes the general form

[tex]y(x)=f(x)[/tex]

whose solution is [tex]f(x)[/tex].

I think expanding means to remove the parentheses/brackets from a problem.

For example: Say we have the expression: 3 (4 + 5). I think expanding means to multiply 3, by every number in the parentheses. So that means:

(3 * 4) + (3 * 5) = 27.

Another way to think about it is to (if you're on paper) draw a line from 3, to all the numbers inside the parentheses. The line that connects from 3 to 4, is signaling for you to multiply 3 * 4 = 12. And the line from 3 to 5 = 3 * 5 = 15. And add them.

Final answer:

In mathematics, 'expand' refers to writing an expression in an extended form using distribution. This can result in a polynomial or an infinite series, as seen in binomial expansion or exponential arithmetic.

Explanation:

In mathematics, to expand means to increase the length of an expression by distributing multiplication over addition or subtraction. For example, expanding (a + b)(c + d) results in ac + ad + bc + bd. This does not change the value of the expression, but rather writes it in an alternative form that might be more useful for further operations, such as simplification or evaluation. Binomial expansion, specifically, refers to expressing a binomial raised to a power as a series of terms, using the binomial theorem, which can sometimes result in an infinite series or a polynomial of finite length. This expansion is applicable in situations like expanding (x + y)^n or when dealing with power series expansions of standard mathematical functions including exponential arithmetic where numbers are expressed as a product of a digit term and an exponential term such as in the notation 4.57 x 10^3.

Answer:

The value is 2401

Step-by-step explanation:

First we need to understand what's means 7 to the 4th power, or 7^4.

7 is the power base, the power base is the number that we are going to repeat the number of times the exponent says.

4 is the exponent, which will tell us how many times to multiply 7.

So , then we would have

1)                7 x 7 x 7 x 7

2)                  49 x 7 x 7

3)                     343 x 7

4)      finally       2401

Answer:

For this case is wrong use the multiplication for P(X=2):

0.42*0.42*0.58*0.58*0.58 = 0.0344

Because we don't take in count the possible nCx ways in order to have the two consumers comfortable, and we are assuming that the first two people are comfortable and the rest is not, and that's not the only possibility. The correct probability for X=2 people comfortable is given by:

[tex]P(X=2)=(5C2)(0.42)^2 (1-0.42)^{5-2}=0.344[/tex]

And as we can see the real answer is 10 times the assumed answer, for this reason is wrong the claim.

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=5, p=0.42)[/tex]

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

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

Where (nCx) means combinatory and it's given by this formula:

[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]

For this case is wrong use the multiplication for P(X=2):

0.42*0.42*0.58*0.58*0.58 = 0.0344

Because we don't take in count the possible nCx ways in order to have the two consumers comfortable, and we are assuming that the first two people are comfortable and the rest is not, and that's not the only possibility. The correct probability for X=2 people comfortable is given by:

[tex]P(X=2)=(5C2)(0.42)^2 (1-0.42)^{5-2}=0.344[/tex]

And as we can see the real answer is 10 times the assumed answer, for this reason is wrong the claim.

Answer:

False

Step-by-step explanation:

The likelihood of an event happening added to its compliment's likelihood must equal 1. The compliment of rolling a number less than 3 on a normal six-sided die is rolling a 3 or more on a normal six-sided die. The probability of rolling 3 or more is 4/6 or 2/3. Adding 1/3 to 2/3 gives us 1 or 100%, thus those events complement each other.

Answer:

Probability that the other side of this coin is a head = 0.5

Step-by-step explanation:

Given that there are two coins: a standard coin with head and tail and a 2 - headed coin.

When tossing a randomly chosen coin once the sample space obtained will be :

Now since we have to find the probability that the other side of this coin which is tossed is also a head which means probability of selecting a 2-headed coin.

From the above cases there are two outcomes which are in favour from total of four outcomes;

Hence, Probability that the other side of this coin is a head = [tex]\frac{1}{2}[/tex] = 0.5 .

Answer:

Step-by-step explanation:

The problem relates to filling 8 vacant positions by either 0 or 1

each position can be filled by 2 ways so no of permutation

= 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2

= 256

b )

Probability of opening of lock in first arbitrary attempt

= 1 / 256

c ) If first fails , there are remaining 255 permutations , so

probability of opening the lock in second arbitrary attempt

= 1 / 255 .

Answer:

x = 14.48

Step-by-step explanation:

first we have to see that we have the measurements from all sides

and we know that the angle between side 21 and 20 is 90 degrees

well to start we have to know the relationships between angles, legs and the hypotenuse.

a: adjacent

o: opposite

h: hypotenuse

sin α = o/h

cos α= a/h

tan α = o/a

let's take the left angle as α

sin α = 21/29

α = sin^-1 (21/29)

α = sin^-1 (0.7241)  

α = 46.397

Now we do the same with the smaller triangle

tan α = o/a

sin 46.397  = x/20

0.724 = x/20

0.724 * 20 = x

14.48 = x

x = 14.48

if we want to check it we can do the same procedure with the other angle

Answer:

Step-by-step explanation:

Hypothesis testing invariably used to test a claim about a population parameter is widely used to check whether they hypothetical claim made is right.

For example, mean scores of a particular college is more than 75% is tested with hypothesis as setting null as equal to 75% and alternate >75%

Processes are done stepwise from the sample collected and conclusion made

Confidence interval on the other hand is the range of values within which the parameter is expected to lie at a certain confidence level

Estimated population parameter is provided for error known as margin of error depending upon the confidence level, and an interval is prepared which guarantees to the extent of confidence that parameters will fall within.

Hypothesis testing can be concluded with the use of confidence intervals also.

Answer:

[tex] \bar X = \frac{\sum_{i=1}^5 x_i f_i}{n} = \frac{2906}{40}= 72.65[/tex]

[tex] s^2 = \frac{213856 -\frac{2906}{40}}{40-1}=70.131[/tex]

[tex] s = \sqrt{70.131}= 8.374[/tex]

Step-by-step explanation:

For this case we can calculate the expected value with the following table"

Class    Midpoint(xi)   Freq. (fi)       xi fi          xi^2 * fi

61-66     63.5                13             825.5      52419.5  

67-72     69.5               10              695        48302.5

73-78     75.5                3              226.5      17100.75

79-84     81.5                11              896.5      73064.75

85-90    87.5                3              262.5       22968.75

________________________________________________

Total                            40            2906          213856

For this case the midpoint is calculated as the average between the minimum and maximum point for each class.  

The expected value can be calculated with the following formula:

[tex] \bar X = \frac{\sum_{i=1}^5 x_i f_i}{n} = \frac{2906}{40}= 72.65[/tex]

For this case n =40 represent the total number of obervations given,  

And for the sample variance we can use the following formula:

[tex] s^2 = \frac{\sum x^2_i f_i -\frac{\sum x_i f_i}{n}}{n-1}[/tex]

And if we replace we got:

[tex] s^2 = \frac{213856 -\frac{2906}{40}}{40-1}=70.131[/tex]

And for the deviation we take the square root:

[tex] s = \sqrt{70.131}= 8.374[/tex]

To calculate the sample standard deviation and sample variance, first calculate the sample mean, then calculate the sample variance, and finally find the square root of the sample variance to get the sample standard deviation.

To calculate the sample standard deviation and sample variance for the given frequency distribution of heart rates, we need to follow these steps:

Using the provided data, the sample standard deviation and sample variance can be calculated as follows:

Sample mean = (65 * 13 + 69.5 * 10 + 75.5 * 3 + 81.5 * 11 + 87.5 * 3) / (13 + 10 + 3 + 11 + 3) ≈ 72.74

Sample variance = [(65 - 72.74)² * 13 + (69.5 - 72.74)² * 10 + (75.5 - 72.74)² * 3 + (81.5 - 72.74)² * 11 + (87.5 - 72.74)² * 3] / (13 + 10 + 3 + 11 + 3 - 1) ≈ 54.21

Sample standard deviation = √(54.21) ≈ 7.36

The Z-score for the 75th percentile is approximately 0.675. The mean insurance cost is $1,650, and the cut off for the 75th percentile is $1,800. The standard deviation of insurance premiums in LA is about $222.22.

(a) The Z-score corresponding to the 75th percentile of the standard normal distribution is approximately 0.675. This is determined referring to standard statistical tables or calculators.

(b) The mean insurance cost is given as $1,650.

The cut off for the 75th percentile is $1,800. This is based on the given information that 25% of California residents pay more than $1,800.

(c) To determine the standard deviation, we first subtract the mean from the 75th percentile value ($1,800 - $1,650 = $150), then divide by the Z-score (150 / 0.675 = approximately $222.22). So, the standard deviation for LA auto insurance premiums is about $222.22.

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The Z-score for the 75th percentile of the standard normal distribution is approximately 0.67. The mean insurance cost is $1,650, and the cutoff for the 75th percentile is $1,800. While we can't identify the standard deviation of insurance premiums in LA without additional data, it would be theoretically possible to find it using the Z-score formula.

In response to your question, let's take it step by step:

(a): The Z-score that corresponds with the 75th percentile of the standard normal distribution is approximately 0.67. You can find this value by utilizing a look-up table (table B.1).

(b): Given in the question, the mean insurance cost for California residents is $1,650.

The 75th percentile (or cutoff) is $1,800, signifying that 25% of residents pay more than this amount.

(c): Unfortunately, the information provided in the question doesn't offer enough data to figure out the standard deviation for insurance premiums in LA directly. However, based on the details given, you can conclude that for a Z-score of 0.67, the corresponding cost is $1,800. Using the Z-score formula, (X - μ) / σ, where X is the value from the dataset, μ is the mean, and σ is the standard deviation, you could theoretically solve for σ (standard deviation) if all other values are known.

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Answer:

Step-by-step explanation:

We are provided that IQs of 16 students from one area of a city had a mean of 107 and a standard deviation of 10 while the IQs of 14 students from another area of the city had a mean of 112 and a standard deviation of 8.

And we have to check that is there a significant difference between the IQs of the two groups.

Firstly let,    Null Hypothesis, [tex]H_0[/tex] : The two groups have same IQs { [tex]\mu_1 = \mu_2[/tex] }

          Alternate Hypothesis, [tex]H_1[/tex] : The two groups have different IQs{ [tex]\mu_1 \neq \mu_2[/tex]}

Since we don't know about population standard deviations;

The test statistics we will use here will be ;

            [tex]\frac{(X_1bar - X_2bar)- (\mu_1 - \mu_2) }{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }[/tex] follows t distribution with [tex](n_1 + n_2 -2)[/tex] degree

                                                    of freedom { [tex]t_n__1 +n_2 - 2[/tex] }

 Here, [tex]X_1bar[/tex] = 107      [tex]X_2bar[/tex] = 112         [tex]s_1[/tex] = 10          [tex]s_2[/tex] = 8

           [tex]n_1[/tex] = 16                [tex]n_2[/tex] = 14

           [tex]s_p[/tex] = [tex]\sqrt{\frac{(n_1 - 1)*s_1^{2} + (n_2 -1)*s_2^{2} }{(n_1 + n_2 -2)} }[/tex] = 9.1261

  Test statistics = [tex]\frac{(107-112) - 0}{9.1261*\sqrt{\frac{1}{16} +\frac{1}{14} } }[/tex]  follows [tex]t_2_8[/tex]

                           = -1.50

Now at 1% level of significance t table is giving the critical value of -2.467 and our test statistics is higher than this means it does not fall in the rejection region so we will accept our null hypothesis and conclude that there is no significant difference between the IQs of the two groups.

This question can be addressed by conducting a two-sample t-test to determine if there is a significant difference between the mean intelligence quotients of students from two areas of a city. The steps include calculating pooled standard deviation, followed by standard error, calculating the t-score, and comparing it to a critical value. You can either reject or fail to reject the null hypothesis based on these results.

The question is asking if there is a significant difference in the mean intelligence quotients (IQs) of students between two areas of a city. Conducting a two-sample t-test can address this. First, let's define the null and alternative hypotheses.

To perform this test, follow these steps:

Remember, depending on the specific results, we may or may not find enough evidence to support the alternative hypothesis.

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Answer:

OPTION C: [tex]$ \textbf{37} \frac{\textbf{28}}{\textbf{8}} $[/tex] pounds.

Step-by-step explanation:

From the figure we can see that there are three dots against [tex]$ 3 \frac{7}{8} $[/tex].

That means it becomes [tex]$ 3 \times 3\frac{7}{8} $[/tex].

Note that if there is a mixed fraction of the form [tex]$ a \frac{b}{c} $[/tex]   =   [tex]$ a + \frac{b}{c} $[/tex].

Therefore, [tex]$ 3 \times 3\frac{7}{8} = 3 \times \bigg(3 + \frac{7}{8} \bigg ) $[/tex]                ... (1)

Similarly, against 4 there are 2 dots.

So, it should be [tex]$ 4 \times 2 $[/tex] pounds.                   ...(2)

3 dots against [tex]$ 4 \frac{1}{8} $[/tex].

So, it becomes [tex]$ 3 \times \bigg(4 + \frac{1}{8} \bigg) $[/tex]                       ...(3)

Similarly, 2 dots against [tex]$ 4 + \frac{2}{8} $[/tex].

This will become [tex]$ 2 \times \bigg( 2 + \frac{2}{8} \bigg) $[/tex]                  ...(4)

Now, to calculate the total pound, we simply add (1), (2), (3) & (4).

⇒    [tex]$ 3 \times \bigg(3 + \frac{7}{8} \bigg ) $[/tex]     [tex]$ + $[/tex]      [tex]$ 4 \times 2 $[/tex]       +     [tex]$ 3 \times \bigg(4 + \frac{1}{8} \bigg) $[/tex]       [tex]$ + $[/tex]        [tex]$ 2 \times \bigg( 2 + \frac{2}{8} \bigg) $[/tex]

⇒    [tex]$ 9 + \frac{21}{8} + 8 + 12 + \frac{3}{8} + 8 + \frac{4}{8} $[/tex]

⇒    [tex]$ \bigg ( 9 + 8 + 12 + 8 \bigg) + \bigg( \frac{21 + 3 + 4}{8} \bigg ) $[/tex]

⇒ [tex]$ \textbf{37} \textbf {+} \frac{\textbf{28}}{\textbf{8}} $[/tex]  [tex]$ \textbf{=} \hspace{1mm} \textbf{37} \frac{\textbf{28}}{\textbf{8}} $[/tex] which is the required answer.

Answer: $115.5

Step-by-step explanation:

The woman bought a coat for $99.95 and some gloves for $7.95. This means that the total amount of money that the woman would have paid for the coat without tax is

99. 95 + 7.95 = $107.9

If the sales tax was 7%, then the amount paid as sales tax would be

7/100 × 107.9 = 0.07 × 107.9 = $7.553

Therefore, the amount that she would pay to purchase the coat and the gloves is

107.9 + 7.553 = 115.453

= $115.5 rounded up to the nearest cent.

The woman spent

[tex]99.95+7.95=107.9[/tex]

dollars in total. Given the 7% sales tax, it means that she actually paid 107% of this amount. So, the final price was

[tex]107.9\cdot\dfrac{107}{100}=1.079\cdot 107=115.453\approx 115.45[/tex]

dollars.

Answer:

a) 4.076

b) 3.9

c) variance for executives=1.128

variance for middle mangers=0.73

d)standard deviation for executives=1.062

standard deviation for middle mangers=0.854

e) Overall job satisfaction for senior executives is higher than middle manager.

Step-by-step explanation:

IS senior executives

Job Satisfaction       1    2          3       4     5

Probability          0.05 0.093 0.03 0.425 0.41

IS middle manager

Job Satisfaction  1       2    3        4    5

Probability         0.01 0.1 0.12 0.46 0.28

Let X denotes IS senior executive and Y denotes IS middle manager.

a)

E(X)=∑x*p(x)=1*0.05+2*0.093+3*0.03+4*0.425+5*0.41

E(X)=0.05+0.186+0.09+1.7+2.05

E(X)=4.076

b)

E(Y)=∑y*p(y)=1*0.1+2*0.1+3*0.12+4*0.46+5*0.28

E(Y)=0.1+0.2+0.36+1.84+1.4

E(Y)=3.9

c)

V(x)=∑x²*p(x)-(∑x*p(x))²

∑x²*p(x)=1*0.05+4*0.093+9*0.03+16*0.425+25*0.41

∑x²*p(x)=0.05+0.372+0.27+6.8+10.25

∑x²*p(x)=17.742

V(x)=17.742-(4.076)²

V(x)=1.128

V(y)=∑y²*p(y)-(∑y*p(y))²

∑y²*p(y)=1*0.1+4*0.1+9*0.12+16*0.46+25*0.28

∑y²*p(y)=0.1+0.4+1.08+7.36+7

∑y²*p(y)=15.94

V(y)=15.94-(3.9)²

V(y)=0.73

d)

S.D(x)=√V(x)

S.D(x)=√1.128

S.D(x)=1.062

S.D(y)=√V(y)

S.D(y)=0.854

e)

Overall job satisfaction for senior executives is more than middle manager as expected value of senior executives is greater than expected value of middle manger with relatively higher variability than middle manager.

Answer:

The correct option is D i.e. Quantitative because numerical values found by either measuring or counting are used to describe the data.

Step-by-step explanation:

As the number of respondents is a numerical value and is identified by counting thus it is a quantitative variable. Also all the other options are incorrect.

A is incorrect because the reason described is not the property of  quantitative data.

B is incorrect because the data is not described in descriptive terms.

C is incorrect because the reason described in not a property of qualitative data.

Answer: i think its B

Step-by-step explanation:

In the given study the response variable is the exam performance, the explanatory variable is the light level, and the blocking variable is the gender of the students.

In this study, the response variable is the exam performance of the students. This is what is being measured as an outcome. The explanatory variable is the light level. The different light levels (fluorescent overhead lighting, yellow overhead lighting, or no overhead light and only desk lamps) constitute the treatments are its levels. The blocking variable in this case is the gender of the students. By making sure that both males and females are equally represented in each treatment, the researcher is controlling for the effect of gender. The levels of this blocking variable are male and female.

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The study measures the effect of light level (explanatory variable) on student exam performance (response variable), with gender as a blocking variable. Lurking variables and study design elements like random assignment and blinding are crucial for maintaining the validity of the results.

a) In the study described, the response variable is the exam performance of the students. This variable will be measured to assess the impact of different lighting conditions on students' ability to perform on an exam.

b) The explanatory variable, or independent variable, is the type of lighting. The levels of this variable are fluorescent overhead lighting, yellow overhead lighting, and no overhead lighting (only desk lamps).

c) The blocking variable is gender. The researcher wants to make sure that both males and females are equally represented in each treatment to test the hypothesis that light levels might affect genders differently. The levels of this variable are simply male and female.

When selecting participants, it is important to consider random assignment to ensure that each treatment group is similar in all respects other than the treatment itself. The idea of dividing participants to drive without distraction and to text and drive could be problematic due to ethical considerations and the introduction of confounding variables.

Lurking variables that could interfere with the study on light levels might include the time of day the exam is taken, students' prior knowledge and preparation levels, or even the difficulty of the exam itself.

Blinding could be used by ensuring that the person measuring exam performance does not know which lighting condition the student was exposed to, thus preventing any bias in the evaluation of the exam performance.

Answer:

147000 J

Step-by-step explanation:

We are given that

Length of chain=L=50 m

Density of chain=[tex]\rho=10kg/m^3[/tex]

We have to find the work done required to wind the chain into the cylinder if a 50 kg block is attached to the end of the chain.

Work done=[tex]\int_{a}^{b}F(y)dy[/tex]

We have F(y)=[tex]\rho g(50-y)dy[/tex]

a=0 and  b=50

[tex]g=9.8m/s^2[/tex]

Using the formula

Work done=[tex]w_1=10\times 9.8\int_{0}^{50}(50-y)dy[/tex]

Where Length of chain is (50-y) has to be lifted.

Work done=[tex]w_1=10\times 9.8[50y-\frac{y^2}{2}]^{50}_{0}[/tex]

By using the formula [tex]\int x^ndx=\frac{x^{n+1}}{n+1}+C[/tex]

Work done=[tex]w_1=10\times 9.8\times (50(50)-\frac{(50)^2}{2})=98\times (2500-1250)=122500 J[/tex]

When the chain is weightless then the work done required to lift the block attached to the 50 m long chain

Again using the formula

Where f(y)=mg

[tex]w_2=\int_{0}^{50}mgdy[/tex]

We have m=50 kg

[tex]w_2=\int_{0}^{50}50\times 9.8 dy=490[y]^{50}_{0}=490\times 50=24500 J[/tex]

The work done required  to wind the chain into the cylinder if a 50 kg block is attached to the end of the chain=[tex]w_1+w_2=122500+24500=147000 J[/tex]

Answer:

[tex] \bar X = \frac{\sum_{i=1}^n X_i f_i}{n}= \frac{1394}{88}=15.8[/tex]

[tex] s^2 = \frac{88(32372) -(1394)^2}{88(88-1}=118.3[/tex]

[tex]Sd(X) = \sqrt{118.273}=10.9[/tex]

Step-by-step explanation:

Previous concepts

In statistics and probability analysis, the expected value "is calculated by multiplying each of the possible outcomes by the likelihood each outcome will occur and then summing all of those values".

The variance of a random variable Var(X) is the expected value of the squared deviation from the mean of X, E(X).

And the standard deviation of a random variable X is just the square root of the variance.  

Solution to the problem

For this case we can calculate the properties required with the following table:

Interval     Mid point (x)     f           x*f         x^2 *f

_________________________________________

1-10               5.5               40        220         1210

11-20             15.5              15        232.5       3603.75

21-30            25.5             23       586.5       14955.75

>31                35.5             10        355          12602.5

________________________________________

Total                                 88        1394         32372

We assume that the mid point for the class >31 is 35.5 using the problem information.

For this case the expected value would be given by:

[tex] \bar X = \frac{\sum_{i=1}^n X_i f_i}{n}= \frac{1394}{88}=15.8[/tex]

The variance owuld be given by this formula"

[tex] s^2 = \frac{n(\sum x^2 f) -(\sum xf)^2}{n(n-1}[/tex]

And if we replace we got:

[tex] s^2 = \frac{88(32372) -(1394)^2}{88(88-1}=118.3[/tex]

The standard deviation would be just the square root of the variance:

[tex]Sd(X) = \sqrt{118.273}=10.9[/tex]

To compute mean age, sample variance, and standard deviation, establish class midpoints, then follow steps of calculating mean, variance, and standard deviation methodologies. Utilize relevant class midpoints, frequencies and population size.

Given the prehistoric Native American family group of 88 members, we can compute the mean age, sample variance, and sample standard deviation using the group's age ranges and sizes.

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Step-by-step explanation:

The median rent for a one-bedroom apartment stands at $3,460 a month.

Also The estimated cost of annual necessities for a single person is $43,581 — or $3,632 a month, making it the most expensive city for single people to settle down in.

And For a family of four, expect to pay about $91,785 a year for necessities — that's $7,649 per month.

For a family of four, expect to pay about $91,785 a year for necessities — that's $7,649 per month.

Although exact data isn't provided, information on related costs such as average salary and gasoline prices suggest that the average cost of living in San Francisco is high.

The average cost of living in San Francisco is significantly higher than the national average. According to Numbeo, San Francisco's overall cost of living index is 176.89, which is 76.89% higher than the U.S. average of 100. This means that you can expect to pay about 77% more for goods and services in San Francisco than you would in the average American city.

However, we can infer that the cost of living is high, considering the mean starting salary for San Jose State University graduates, nearby to San Francisco, is at least $100,000 per year. This suggests that a significant income is required to support oneself in the Bay Area.

Other factors indirectly hint at the costs associated with San Francisco living. For instance, the average cost of unleaded gasoline in the Bay Area was once $4.59, which is notably high. These pieces of information, though incomplete, indicate a high cost of living.

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Answer:

Work done will be equal to 5.2059 lb-ft

Step-by-step explanation:

We have given force F = 10 lb

Spring is stretched to 2 in

So x = 2 in

As 1 inch = 0.0833 feet

So 2 inch = 2×0.0833 = 0.1666 feet

From hook's law we know that F = Kx , here K is spring constant and x is spring elongation

So [tex]10=K\times 0.1666[/tex]

K = 60.024 lb/feet

Now new elongation x = 5 in

So 5 in = 5×0.0833 = 0.4165 feet

Work done is given by [tex]W=\frac{1}{2}Kx^2[/tex]

So [tex]W=\frac{1}{2}\times 60.02\times 0.4165^2=5.205lb-ft[/tex]

So work done will be equal to 5.2059 lb-ft

Answer:

a)  P(2)=0.270

b) P(X>3)=0.605

c)  P=0.410

Step-by-step explanation:

We know that customers arrive at a grocery store at an average of 2.1 per minute. We use the  Poisson distribution:

[tex]\boxed{P(k)=\frac{\lambda^k \cdot e^{-\lambda}}{k!}}[/tex]

a)  In this case: [tex]\lambda=2.1[/tex]

[tex]P(2)=\frac{2.1^2 \cdot e^{-2.1}}{2}\\\\P(2)=0.270[/tex]

Therefore, the probability is P(2)=0.270.

b)  In this case: [tex]\lambda=2\cdot 2.1=4.2[/tex]

[tex]P(X>3)=1-P(X\leq 3)\\\\P(X>3)=1-\sum_{x=0}^3 \frac{4.2^x \cdot e^{-4.2}}{x!}\\\\P(X>3)=1-0.395\\\\P(X>3)=0.605[/tex]

Therefore, the probability is P(X>3)=0.605.

c)  We know that two customers came in in the first minute. That is why we calculate the probability of at least 5 customers entering the other 2 minutes.

In this case: [tex]\lambda=2\cdot 2.1=4.2[/tex]

[tex]P(X\geq 5)=1-P(X<5)\\\\P(X\geq 5)=1-P(X\leq 4)\\\\P(X\geq 5)=1-\sum_{x=0}^4 \frac{4.2^x\cdot e^{-4.2}}{x!}\\\\P(X\geq 5)=1-0.590\\\\P(X\geq 5)=0.410[/tex]

Therefore, the probability is P=0.410.

Suppose X indicates the number of customers that enter a grocery store within one minute.

[tex]\to X \sim \text{Poisson}(2.1)[/tex]

X's probability mass function:

[tex]\to P(X=x)=\frac{e^{-2.1} \times 2.1^{x}}{x!} ,x =0,1,2,..[/tex]

For question 1:

The likelihood of exactly two consumers arriving in a minute:

[tex]P(X=2)=\frac{e^{-2.1} \times 2.1^{2}}{2!}= 0.270016 \approx 0.270[/tex]

For question 2:

Suppose Y represents the average group of consumers who enter a grocery shop in a two-minute period.

[tex]Y \sim \text{Poisson}(2.1\times 2) \ or\ Y \sim \text{Poisson}(4.2)[/tex]

Y's probability mass function:

[tex]P(Y = y) = \frac{e^{-4.2} \times 4.2^{y}}{y!}, y =0,1,2..[/tex]

This is possible that more than three clients will arrive within a two-minute timeframe.

[tex]= P(Y > 3) \\\\= 1 - P(Y \leq 3)\\\\=1- \Sigma^{3}_{y=0} \frac{e^{-4.2} \times 4.2^y}{y!}\\\\=1-0.395403\\\\= 0.604597\approx \ 0.605[/tex]

For question 3:

Given that exactly two customers arrive in the first minute, the likelihood of at least seven clients arriving in three minutes is  

[tex]= \text{P( at least 5 customers arrive in two minutes)} \\\\= P(Y \geq 5)\\\\= 1 - P(Y<5)\\\\=1-P(Y \leq 4)\\\\= 1 - 0.589827\\\\= 0.410173 \approx \ 0.410[/tex]

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