The Consumer Reports National Research Center conducted a telephone survey of 2000 adults to learn about the major economic concerns for the future (Consumer Reports, January 2009). The survey results showed that 1760 of the respondents think the future health of Social Security is a major economic concern.

a. What is the point estimate of the population proportion of adults who think the future health of Social Security is a major economic concern (to 2 decimals)?
b. At 90% confidence, what is the margin of error (to 4 decimals)?
c. Develop a 90% confidence interval for the population proportion of adults who think the future health of Social Security is a major economic concern (to 3 decimals).(, )
d. Develop a 95% confidence interval for this population proportion (to 4 decimals).

Answer:

option 4

The value is not unusual.

Step-by-step explanation:

If the probability of revenue between $747.89 and $818.68 million is low i.e. less than 5% then we can say that it would be unusual for a randomly selected company to have a revenue between $747.89 and 818.68 million.

We are given that revenue found to follow a bell curve. Also, we are given that mean=815.425 and standard deviation=22.148.

P(747.89<x<818.68)=?

P((747.89-815.425)/22.148<z<(818.68-815.425)/22.148)=P(-3.05<z<0.15)

P(747.89<x<818.68)=0.4989+.0596=0.5585

Thus, the probability for a randomly selected company to have a revenue between $747.89 and 818.68 million is not low and so the value is not unusual.

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:

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: d, p = 0.4493

Step-by-step explanation: this question is solved using a possion probability distribution because the event is occurring at a fixed rate.

For this question of ours the fixed rate is the fact that 12 customers visiting the shop within 15 minutes.

For this question our fixed rate (u) = 12/15 = 0.8

The probability distribution for possion is given as

P(x=r) = (e^-u * u^r) / r!

At this point x = 0 ( no customers coming to the shop)

P(x=0) =( e^-0.8 * 0.8^0)/ 0!

P(x=0) = (e^-0.8 * 1)/1

P(x=0) = e^-0.8

P(x=0) = 0.4493

To maximize your return, you should sell the house after approximately 4.3 years.

To maximize your return, you should sell the house when its value plus the accumulated interest in the fund is the highest. Let's calculate the number of years for this to happen:

Value of the house after y years = $350,000 + $43,750y

Value of the fund after y years = $350,000(1 + 0.07/52)^(52y)

To find the number of years that maximizes the return, we need to find the point where the two values are equal. This can be done by solving the equation:

$350,000 + $43,750y = $350,000(1 + 0.07/52)^(52y)

Solving this equation will give us the number of years to maximize the return.

After calculating the equation, we find that after approximately 4.3 years, you should sell the house to maximize your return.

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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 :- y = 9/2

and x = 1/2

Answer:

Step-by-step explanation:

The given system of equations is expressed as

4x + 2y = 16 - - - - - - - - - - - - -1

3x + 3y =15- - - - - - - - - - - - - -2

We would eliminate x by multiplying equation 1 by 3 and equation 2 by 4. It becomes

12x + 6y = 48 - - - - - - - - - - - -3

12x + 12y = 60 - - - - - - - - - - - -4

Subtracting equation 4 from equation 3, it becomes

- 6y = - 12

Dividing the left hand side and the right hand side of the equation by - 6, it becomes

- 6y/-6 = - 12/ - 6

y = 2

Substituting y = 2 into equation 1, it becomes

4x + 2 × 2 = 16

4x + 4 = 16

Subtracting 4 from the left hand side and the right hand side of the equation, it becomes

4x + 4 - 4 = 16 - 4

4x = 12

Dividing the left hand side and the right hand side of the equation by 4, it becomes

4x/4 = 12/4

x = 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:

[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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Answer:

$9,979,032,100 or $9.979 billion

Step-by-step explanation:

The approximate weight of a $100 bill is 1 gram.

All of the calculations bellow assume that the volume of the bills would not be an issue and only concerns weight.

1 pound is equivalent to approximately 453.59237 grams.

The weight in grams that the Big Falcon Rocket can carry is:

[tex]W= 220,000*453.59237=99,790,321.4\ g[/tex]

Since each bill weighs 1 gram, the number of bills it could carry, rounded to nearest whole bill is 99,790,321. The total amount it could carry is:

[tex]A=99,790,321*\$100=\$9,979,032,100[/tex]

Answer:

(A) The minimum sample size required achieve the margin of error of 0.04 is 601.

(B) The minimum sample size required achieve a margin of error of 0.02 is 2401.

Step-by-step explanation:

Let us assume that the percentage of support for the candidate, among voters in her district, is 50%.

(A)

The margin of error, MOE = 0.04.

The formula for margin of error is:

[tex]MOE=z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}}[/tex]

The critical value of z for 95% confidence interval is: [tex]z_{\alpha/2}=1.96[/tex]

Compute the minimum sample size required as follows:

[tex]MOE=z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}}\\0.04=1.96\times \sqrt{\frac{0.50(1-0.50)}{n}}\\(\frac{0.04}{1.96})^{2} =\frac{0.50(1-0.50)}{n}\\n=600.25\approx 601[/tex]

Thus, the minimum sample size required achieve the margin of error of 0.04 is 601.

(B)

The margin of error, MOE = 0.02.

The formula for margin of error is:

[tex]MOE=z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}}[/tex]

The critical value of z for 95% confidence interval is: [tex]z_{\alpha/2}=1.96[/tex]

Compute the minimum sample size required as follows:

[tex]MOE=z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}}\\0.02=1.96\times \sqrt{\frac{0.50(1-0.50)}{n}}\\(\frac{0.02}{1.96})^{2} =\frac{0.50(1-0.50)}{n}\\n=2401.00\approx 2401[/tex]

Thus, the minimum sample size required achieve a margin of error of 0.02 is 2401.

Final answer:

It took Mark 26 minutes to run from mile marker 1 to mile marker 4. His average speed was 0.5038 miles per minute for that segment. To complete the race in 110 minutes, Mark's average speed from mile marker 9 to the finish line must be 0.4415 miles per minute.

Explanation:

To find how long it took Mark to run from mile marker 1 to mile marker 4, we need to add the times it took him to run from mile marker 1 to mile marker 2 and from mile marker 2 to mile marker 4. Mark took 6 minutes to run from mile marker 1 to mile marker 2, and 20 minutes to run from mile marker 2 to mile marker 4. So the total time it took him to run from mile marker 1 to mile marker 4 is 6 minutes + 20 minutes = 26 minutes.

To find Mark's average speed as he ran from mile marker 1 to mile marker 4, we need to divide the total distance he ran by the total time it took him. The total distance from mile marker 1 to mile marker 4 is 13.1 miles. So the average speed is 13.1 miles / 26 minutes = 0.5038 miles per minute.

To complete the race in 110 minutes, Mark has 39 minutes left after passing mile marker 9. To find his average speed as he travels from mile marker 9 to the finish line, we need to divide the remaining distance by the remaining time. The remaining distance is 26.22 miles - 9 miles = 17.22 miles. So his average speed is 17.22 miles / 39 minutes = 0.4415 miles per minute.

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:

(c)approximately Normal, mean 112, standard deviation 1.414.

Step-by-step explanation:

To solve this problem, we have to understand the Central Limit Theorem

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], a large sample size can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\frac{\sigma}{\sqrt{n}}[/tex].

In this problem, we have that:

[tex]\mu = 112, \sigma = 20, n = 200[/tex]

Using the Central Limit Theorem

The distribution of the sample mean IQ is approximately Normal.

With mean 112

With standard deviation [tex]s = \frac{20}{\sqrt{200}} = 1.414[/tex]

So the correct answer is:

(c)approximately Normal, mean 112, standard deviation 1.414.

The distribution of sample means for 200 adults is approximately normal due to the Central Limit Theorem, with a mean of 112 and a standard deviation of 1.414, making option (e) the correct choice.

The question asks about the distribution of the sample mean IQ of 200 randomly selected adults from a large population where the mean IQ is 112 with a standard deviation of 20. According to the Central Limit Theorem, when the sample size is large, the distribution of the sample means will be approximately normal (normal by the Central Limit Theorem), even if the source population itself is not perfectly normal.

The mean of the sampling distribution of the sample mean will be the same as the mean of the population, so the mean will be 112.

However, the standard deviation of the sampling distribution (often called the standard error) is equal to the standard deviation of the population divided by the square root of the sample size. So, the standard deviation for the sample mean for 200 adults would be 20 / √200, which is about 1.414. Therefore, the correct choice is (e) exactly Normal, mean 112, standard deviation 1.414.

Answer:

Part a: The Future value of the annuity after 40 years is $518113.24.

Part b: The per year withdrawal in retirement for 25 years will be $48536.19.

Step-by-step explanation:

As the numbers are appearing as a duplication taking all these values as single.

Part a

Future value is given as

[tex]FV=PMT \times [\frac{{(1+I)}^{N}-1}{I}][/tex]

Here

[tex]FV=PMT \times [\frac{{(1+I)}^{N}-1}{I}]\\\\FV=2000 \times [\frac{({1+.08})^{40}-1}{.08}]\\FV=\$ 518113.03[/tex]

So the Future value of the annuity after 40 years is $518113.24.

Part b

Per year withdrawal is given as

[tex]PY=\frac{Value}{\frac{1 - \frac{1}{(1+I)^N}}{I}}[/tex]

Here

So the value is given as

[tex]PY=\frac{Value}{\frac{1 - \frac{1}{(1+I)^N}}{I}}\\PY=\frac{518113}{\frac{1 - \frac{1}{(1+0.08)^{25}}}{0.08}}\\PY=\frac{518113}{10.6747}\\PY=\$ 48536.19[/tex]

So the per year withdrawal in retirement for 25 years will be $48536.

To calculate the future value of the social security annuity with an 8% interest rate over 40 years, we use the future value of an annuity formula. For withdrawals, we then use the future value as the principal to determine yearly withdrawals during a 25-year retirement at the same interest rate.

The question involves calculating the future value of an annuity and determining the equivalent yearly withdrawals during retirement. For part (a), we will use the future value of annuity formula to find the future worth of the $2,000 annual payment into the social security system given an interest rate of 8% over 40 years. Part (b) involves finding the annuity payment given a fixed principal (the future value calculated in part a), an interest rate of 8%, and a withdrawal period of 25 years (the expected span of retirement).

Calculation for part (a):

The future value FV of an annuity can be calculated using the formula FV = P * [((1 + r)^n - 1) / r], where P is the payment amount per period, r is the interest rate, and n is the total number of payments. In this scenario, P = $2,000, r = 0.08 (8%), and n = 40. Plugging the numbers into the formula will give us the future value of the social security savings.

Calculation for part (b):

For withdrawals during retirement, we start with the future value calculated in part (a) as the principal amount to be entirely withdrawn over 25 years. The annuity payment A can be found using the formula A = P * [r / (1 - (1 + r)^(-n))], where P is the future value from part (a), r is again 0.08, and n is 25. This will provide the yearly withdrawal amount that can be made during retirement.

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.

The entries that will be interpreted as numbers are a) 1.56e-9, b) -4.99, d) 1.9435, and g) 3.25E4. Entries c), e), f), h), and i) violate the restrictions mentioned.

The entries that will be interpreted as numbers are:

Entries a), b), d), and g) are written in proper scientific notation and do not violate any of the restrictions mentioned. Entry c) violates the rule of substituting the letter O for zero or the letter l for 1. Entry e) has a space between the number and the exponent, violating the rule of not having a space in a number. Entry f) includes a dollar sign in the number, which is not allowed. Entry h) has a comma, violating the rule of not having commas in numbers. Entry i) includes the units 'inches', which is not allowed.

Final answer:

Entries 1.56[tex]e^{-9[/tex], -4.99, and 1.9435 are valid numbers that would be interpreted by WebAssign, while 3.25E4 is also correct as it is proper scientific notation. (Option a, b,d,g)

Explanation:

When entering numbers in WebAssign, it is important to use the correct format to ensure that the numbers are interpreted accurately. Here are which of the entries will be interpreted as numbers:

a) 1.56[tex]e^{-9[/tex]: This is a correct representation of a number in scientific notation.

b) -4.99: This is a valid negative number.

c) 40O0: This contains a letter and would not be interpreted as a number.

d) 1.9435: This is a simple decimal number.

e) 1.56 [tex]e^{-9[/tex]: This is not correctly formatted for scientific notation due to the space.

f) $2.59: This includes a dollar sign, which is not valid for a number entry.

g) 3.25E4: This is correctly written in scientific notation.

h) 5,000: Commas are not allowed in number entries.

i) 1.23 inches: This includes units, which should not be included in a number entry.

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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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:

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:

Taylor = 50%

Moore = 25%

Jenkins = 25%

Step-by-step explanation:

Assuming there are no other candidates and that someone has to win the election, the probabilities of Taylor, Moore, and Jenkins winning the election must add up to 1 or 100%.

[tex]T+M+J = 1\\[/tex]

Since Moore and Jenkins have about the same probability of winning, and Taylor is believed to be twice as likely to win as either of the others:

[tex]M=J\\T=2J\\J+J+2J=1\\J=0.25\\M=J=0.25\\T=2*0.25=0.5[/tex]

Taylor has a probability of 50% of winning the election.

Moore has a probability of 25% of winning the election.

Jenkins has a probability of 25% of winning the election.

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:

Total = N(1) +N(2)+N(3) = 10+14+200 = 224 sequences

224 sequences would NOT be permitted.

Step-by-step explanation:

1) all four digits identical

Number of possible PIN with all four digits identical N(1) = 1×1×1×10 = 10 possible PIN

2) sequence of consecutive ascending or descending digits N(2) = 14

Only Ascending starting with 0,1,2 = 3

only descending for 7,8,9 = 3

and both for 3,4,5,6 = (2×4)

Total = 3+3+(2×4) = 14

3) any sequence starting with 19 or 20 (birth years

N(3) = 2 × 1×1×10×10 = 2×100 = 200

N(3) = 200

(Note : 100 possible ways for each of 19 or 20s)

Total = N(1) +N(2)+N(3) = 10+14+200 = 224

224 sequences would NOT be permitted.

[tex]\boxed{t=11.3h}[/tex]

In this problem we have the following data:

[tex]r:speed \\ \\ d:distance \\ \\ t:time \\ \\ \\ r=33.2m/h \\ \\ d=375.16m \\ \\ \\ So \ our \ goal \ is \ to \ find \ t[/tex]

Speed, time and distance are related with the following formula:

[tex]r=\frac{d}{t} \\ \\ \\ Solving \ for \ t: \\ \\ t=\frac{d}{r} \\ \\ t=\frac{375.16}{33.2} \\ \\ \boxed{t=11.3h}[/tex]

Speed: https://brainly.com/question/14095729

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(a) 0.611, (b) 0.218, (c) 0.611. Probability of all good, exactly two good, and at least two good units, respectively.

To solve these probability problems, we can use combinations. Let's calculate:

Given:

Total number of microwave ovens = 12

Number of defective units = 3

Number of good units = Total - Defective = 12 - 3 = 9

(a) Probability that all four units are good:

The probability of selecting one good unit = (Number of good units) / (Total number of units) = 9/12

For all four units to be good, we need to multiply this probability four times since each selection is independent:

P(all four are good) = (9/12)×  (9/12)  (9/12)× (9/12) = (9/12)^4

(b) Probability that exactly two units are good:

To find this, we need to choose 2 good units out of 4 and 2 defective units out of the remaining 12 - 4 = 8 units.

Number of ways to choose 2 good units out of 4 = C(4, 2) = 6

Number of ways to choose 2 defective units out of 8 = C(8, 2) = 28

Total number of favorable outcomes = 6 × 28

Total possible outcomes = C(12, 4)

So, the probability is:

P(exactly two units are good) = (6 ×28) / C(12, 4)

(c) Probability that at least two units are good:

This includes the cases where exactly two, three, or four units are good. We've already calculated the probability for exactly two units being good.

Probability that exactly three units are good:

Similar to the previous case, calculate the combinations for choosing 3 good units out of 4 and 1 defective unit out of the remaining 8.

Number of ways to choose 3 good units out of 4 = C(4, 3) = 4

Number of ways to choose 1 defective unit out of 8 = C(8, 1) = 8

Total number of favorable outcomes for exactly three units being good = 4 ×8

Now, the probability for at least two units being good is the sum of probabilities for exactly two, three, and four units being good:

P(at least two units are good) = P(exactly two units are good) + P(exactly three units are good) + P(all four units are good)

Sure, let's calculate each probability:

(a) Probability that all four units are good:

[tex]\[ P(\text{all four are good}) = \left(\frac{9}{12}\right)^4 = \left(\frac{3}{4}\right)^4 = \frac{81}{256} \][/tex]

(b) Probability that exactly two units are good:

[tex]\[ \text{Number of ways to choose 2 good units out of 4} = C(4, 2) = \frac{4!}{2! \times (4-2)!} = 6 \][/tex]

Number of ways to choose 2 defective units out of 8 =

[tex]C(8, 2) = \frac{8!}{2! \times (8-2)!} = 28 \][/tex]

Total possible outcomes:  [tex]\( C(12, 4) = \frac{12!}{4! \times (12-4)!} = 495 \)[/tex]

[tex]\[ P(\text{exactly two units are good}) = \frac{6 \times 28}{495} = \frac{168}{495} = \frac{56}{165} \][/tex]

(c) Probability that at least two units are good:

[tex]\[ P(\text{at least two units are good})[/tex]= [tex]P(\text{exactly two units are good}) + P(\text{exactly three units are good}) + P(\text{all four units are good}) \][/tex]

Now, let's calculate the probability for exactly three units being good:

[tex]\[ \text{Number of ways to choose 3 good units out of 4} = C(4, 3) = \frac{4!}{3! \times (4-3)!} = 4 \][/tex]

Number of ways to choose 1 defective unit out of 8

[tex]= C(8, 1) = \frac{8!}{1! \times (8-1)!} = 8 \][/tex]

[tex]\[ P(\text{exactly three units are good}) = \frac{4 \times 8}{495} = \frac{32}{495} \][/tex]

Now, we can calculate [tex]\( P(\text{at least two units are good}) \):[/tex]

[tex]\[ P(\text{at least two units are good}) = \frac{56}{165} + \frac{32}{495} + \frac{81}{256} \][/tex]

[tex]\[ P(\text{at least two units are good}) = \frac{56}{165} + \frac{32}{495} + \frac{81}{256} \]\[ = \frac{4480}{12870} + \frac{1056}{12870} + \frac{3165}{12870} \]\[ = \frac{4480 + 1056 + 3165}{12870} \]\[ = \frac{8696}{12870} \][/tex]

[tex]\[ \approx 0.675 \][/tex]

So, the probabilities are:

[tex]a) Probability that all four units are good: \( \frac{81}{256} \)(b) Probability that exactly two units are good: \( \frac{56}{165} \)(c) Probability that at least two units are good: \( \frac{8696}{12870} \)[/tex]

Answer: he have 43.6 miles to his destination after 51 minutes of driving.

Step-by-step explanation:

let x represent his total driving time (in minutes).

Let y represent the number of miles that Ryan has to his destination.

If we plot y on the vertical axis and x on the horizontal axis, a straight line would be formed. The slope of the straight line would be

Slope, m = (58 - 46.8)/(33 - 47)

m = 11.2/- 14 = - 0.8

The equation of the straight line can be represented in the slope-intercept form, y = mx + c

Where

c = intercept

m = slope

To determine the intercept, we would substitute x = 33, y = 58 and m = - 0.8 into y = mx + c. It becomes

58 = - 0.8 × 33 + c

58 = - 26.4 + c

c = 58 + 26.4

c = 84.4

The linear function becomes

y = - 0.8x + 84.4

Therefore, after 51 minutes, the number of miles that he has to his destination is

y = - 0.8 × 51 + 84.4

y = 43.6

Answer:

The middle 60 students fall between 63.48 inches and 68.52 inches.

Step-by-step explanation:

Problems of normally distributed samples can be 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 = 66, \sigma = 3[/tex]

Between which two heights (in inches) do the middle 60 students fall?

The normal probability distribution is symmetric. So the middle 60% fall from a pvalue of 0.50 - 0.60/2 = 0.20(lower bound) to a pvalue of 0.50 + 0.60/2 = 0.80(upper bound)

Lower bound

X when Z has a pvalue of 0.20.

So X when [tex]Z = -0.84[/tex]

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]-0.84 = \frac{X - 66}{3}[/tex]

[tex]X - 66 = -0.84*3[/tex]

[tex]X = 63.48[/tex]

Upper bound

X when Z has a pvalue of 0.80.

So X when [tex]Z = 0.84[/tex]

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]0.84 = \frac{X - 66}{3}[/tex]

[tex]X - 66 = 0.84*3[/tex]

[tex]X = 68.52[/tex]

The middle 60 students fall between 63.48 inches and 68.52 inches.

Final answer:

The middle 60 students in a class with normally distributed heights, a mean of 66 inches, and a standard deviation of 3 inches, fall between heights of 63.5 and 68.5 inches.

Explanation:

The student has asked about finding the range of heights for the middle 60 students in a class of 100, where the heights are normally distributed with a mean of 66 inches and a standard deviation of 3 inches. To determine this, we use the concept of percentiles in a normal distribution. Since the middle 60 students represent the 20th to 80th percentile, we need to calculate the heights corresponding to these percentiles. The Z-value associated with the 20th percentile is approximately -0.84, and for the 80th percentile, it's approximately 0.84. Using the Z-score formula Z = (X - mean) / standard deviation, we can solve for X:

For the 20th percentile: X = -0.84 x 3 + 66 = 63.5 inches

For the 80th percentile: X = 0.84 x 3 + 66 = 68.5 inches

Therefore, the middle 60 students have heights ranging from 63.5 to 68.5 inches.

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:

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.