For f(x) = 9x and g(x) = x + 3, find the following functions.

a. (f o g)(x)​;
b. (g o f )(x)​;
c. (f o g )(2)​;
d. (g o f )(2)

Answer:

In 10 seconds, after 75 kernels have popped, they should pop 161/36 kernels.

Step-by-step explanation:

You can solve this problem with a rule of three, if you remove 75 kernels, you have 115 left, and in 10 seconds you may expect a proportional amount of popcorn kernels dropping.

If for 180 kernels 7 drop, then for 115 kernels the amount that drops is

7/180 * 115 = 161/36 popcorn kernels.

Answer:

a. 0.0312 or 3.12%

b. 0.024 or 2.40%

Step-by-step explanation:

a. The probability that randomly selected child has at least one food allergy and a history of severe reaction is determined by the probability of having a food allergy (8%) multiplied by the probability of having a history of severe reaction (39%):

[tex]P = 0.08*0.39\\P=0.0312 = 3.12\%[/tex]

b. The probability that randomly selected child  is allergic to multiple foods is determined by the probability of having a food allergy (8%) multiplied by the probability of being allergic to multiple foods (30%):

[tex]P = 0.08*0.30\\P=0.024 = 2.40\%[/tex]

Answer:

The scores that are less than or equal to 10.8 are considered significantly low.

The scores that are greater than or equal to 32.4 are considered significantly high.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 21.6

Standard Deviation, σ = 5.4

We are given that the distribution of score is a bell shaped distribution that is a normal distribution.

Formula:

[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]

Significantly low score:

[tex]z \leq -2\\z = \displaystyle\frac{x-21.6}{5.4} \leq -2\\\\\displaystyle\frac{x-21.6}{5.4} \leq -2\\\\x\leq -2(5.4) + 21.6\\\Rightarrow x \leq 10.8[/tex]

Thus, scores that are less than or equal to 10.8 are considered significantly low.

Significantly high score:

[tex]z \geq 2\\z = \displaystyle\frac{x-21.6}{5.4} \geq 2\\\\\displaystyle\frac{x-21.6}{5.4} \geq 2\\\\x\geq 2(5.4) + 21.6\\\Rightarrow x \geq 32.4[/tex]

Thus, scores that are greater than or equal to 32.4 are considered significantly high.

Answer:

The probability is [tex]\frac{1}{4}[/tex] or 25%

Step-by-step explanation:

The question states the total number of vehicles, as well as the number of damaged vehicles on a yearly basis. If 50 vehicles in every 200 vehicles per year are damaged, then we can obtain:

Probability of a damaged vehicle in any given year = [tex]\frac{Number of Damaged Vehicles}{Total Number of Vehicles}[/tex]

= [tex]\frac{50}{200}[/tex] = [tex]\frac{1}{4}[/tex] or 25%

Answer:

dA(r) = 6.9 cm²

Step-by-step explanation:

Area of a rectangle is

A(r) = L * W      (1)

Where L stands for length and W for width.

Taking differentials on both sides f equation (1)

dA(r)  =  W*dL  + L*dW

As error in measurments are at most 0,1 cm ( in each  side), then the maximum error in calculating the area is

dA(r) =  24* ( 0,1) (cm²) + 45*(0,1)(cm²) ⇒    dA(r) =  2.4 + 4.5 (cm²)

dA(r) = 6.9 cm²

Answer:

(a) The usual load is not 13 credits.

(b) The probability that a a student at this college takes 16 or more credits is 0.1093.

Step-by-step explanation:

According to the Central limit theorem, if a large sample (n ≥ 30) is selected from an unknown population then the sampling distribution of sample mean follows a Normal distribution.

The information provided is:

[tex]Min.=8\\Q_{1}=13\\Median=14\\Mean=13.65\\SD=1.91\\Q_{3}=15\\Max.=18[/tex]

The sample size is, n = 100.

The sample size is large enough for estimating the population mean from the sample mean and the population standard deviation from the sample standard deviation.

So,

[tex]\mu_{\bar x}=\bar x=13.65\\SE=\frac{s}{\sqrt{n}}=\frac{1.91}{\sqrt{100}}=0.191[/tex]

(a)

The null hypothesis is:

H₀: The usual load is 13 credits, i.e. μ = 13.

Assume that the significance level of the test is, α = 0.05.

Construct a (1 - α) % confidence interval for population mean to check the claim.

The (1 - α) % confidence interval for population mean is given by:

[tex]CI=\bar x\pm z_{\alpha/2}\times SE[/tex]

For 5% level of significance the two tailed critical value of z is:

[tex]z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96[/tex]

Construct the 95% confidence interval as follows:

[tex]CI=\bar x\pm z_{\alpha/2}\times SE\\=13.65\pm (1.96\times0.191)\\=13.65\pm0.3744\\=(13.2756, 14.0244)\\=(13.28, 14.02)[/tex]

As the null value, μ = 13 is not included in the 95% confidence interval the null hypothesis will be rejected.

Thus, it can be concluded that the usual load is not 13 credits.

(b)

Compute the probability that a a student at this college takes 16 or more credits as follows:

[tex]P(X\geq 16)=P(\frac{X-\mu}{\sigma}\geq \frac{16-13.65}{1.91})\\=P(Z>1.23)\\=1-P(Z<1.23)\\=1-0.8907\\=0.1093[/tex]

Thus, the probability that a a student at this college takes 16 or more credits is 0.1093.

Answer:

(a) We reject the null hypothesis that usual load is 13 credits.

(b) Probability that student at this college takes 16 or more credits = 0.10935

Step-by-step explanation:

We are given that the histogram below shows the distribution of the number of credits taken by these students;

 Min   Q1    Median     Mean      SD     Q3      Max

   8     13         14           13.65     1.91      15        18

Also, the counselor decides to randomly sample 100 students by using the registrar's database of students, i.e., n = 100.

So, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu[/tex] = 13 {means that the usual load is 13 credits}

Alternate Hypothesis, [tex]H_1[/tex] : [tex]\mu\neq[/tex] 13 {means that the usual load is not 13 credits}

The test statistics we will use here is ;

              T.S. = [tex]\frac{Xbar - \mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]

where, Xbar = sample mean = 13.65

               s = sample standard deviation = 1.91

               n = sample size = 100

So, test statistics = [tex]\frac{13.65 - 13}{\frac{1.91}{\sqrt{100} } }[/tex] ~ [tex]t_9_9[/tex]

                            = 3.403

Now, since significance level is not given to us so we assume it to be 5%.

At 5% significance level, the t tables gives critical value of 1.987 at 99 degree of freedom. Since our test statistics is more than the critical value which means our test statistics will lie in the rejection region, so we have sufficient evidence to reject null hypothesis.

Therefore, we conclude that the usual load is not 13 credits.

(b) Let X = credits of students

The z score probability distribution is given by;

         Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)

So, probability that student at this college takes 16 or more credits =     P(X [tex]\geq[/tex] 16)

P(X [tex]\geq[/tex] 16) = P( [tex]\frac{X-\mu}{\sigma}[/tex] [tex]\geq[/tex] [tex]\frac{16-13.65}{1.91}[/tex] ) = P(Z [tex]\geq[/tex] 1.23) = 1 - P(Z < 1.23)

                                                 = 1 - 0.89065 = 0.10935 or 11%

Therefore, probability that student at this college takes 16 or more credits is 0.10935.

Answer:

The total number of possible combinations of species that you captured is 11,175.

Step-by-step explanation:

The order that you captured the species is not important. For example, capturing a bird of specie A then a bird of specie B is the same outcomes as capturing a bird of specie B then a bird of specie A. So the combinations formula is used to solve this question.

Combinations formula:

[tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.

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

In this problem, we have that:

Combinations of 2 species from a set of 150. So

[tex]C_{150,2} = \frac{150!}{2!148!} = 11,175[/tex]

The total number of possible combinations of species that you captured is 11,175.

Answer:

[tex]n^{2} - 20n -96 = 0[/tex]

use product and sum method

product = -96

sum = -20

numbers needed = ( -24 , 4)

n - 24 = 0

n + 4 = 0

hence n = 24 and n = -4

[tex]x^{2} + 12 x = 48[/tex]

in the form [tex]ax^{2} +bx +c = 0[/tex]

= [tex]x^{2} +12x - 48[/tex]

make use of the formula :

[tex]\frac{-b+-\sqrt{b^{2} -4ac} }{2a}[/tex]

replace values to make 2 equations :

1.[tex]\frac{-12+\sqrt{12^{2} -4*1*-48} }{2*1}[/tex] = 3.17

2.[tex]\frac{-12-\sqrt{12^{2} -4*1*-48} }{2*1}[/tex] = -15.2

hence x = 3.17 and x = -15.2

[tex]x^{2} -14x+40=0[/tex]

use product and sum method

product = 40

sum = -14

numbers needed = (-10 , -4)

x - 10 = 0

x - 4 = 0

hence x = 10 and x = 4

[tex]5b^{2} -20b-18 = 7[/tex]

in the form [tex]ax^{2} +bx +c = 0[/tex]

this becomes [tex]5b^{2} -20b-18-7[/tex]

= [tex]5b^{2} -20b-25[/tex]

can simplify by 5

= [tex]b^{2} -4b-5 =0\\[/tex]

use product and sum method

product = -5

sum = -4

numbers needed (-5 , 1)

b-5 = 0

b + 1 = 0

hence b = 5 and b = -1

Answer:

Step-by-step explanation:

1) n² - 20n - 96 = 0

n² - 20n + (- 20/2)² = 96 + (- 20/2)²

(n - 10)² = 96 + 100

(n - 10)² = 196

Taking square root of both sides

n - 10 = √196 = 14

n = 14 + 10

n = 24

2) x² + 12x = 48

x² + 12x + (12/2)² = 48 + (12/2)²

(x + 6)² = 48 + 36 = 84

Taking square root of both sides,

x + 6 = 9.2

x = 9.2 - 6

x = 3.2

3) x² - 14x + 40 = 0

x² - 14x = - 40

x² - 14x + (- 14/2)² = - 40 + (- 14/2)²

(x - 7)² = - 40 + 49 = 9

Taking square root of both sides,

x - 7 = 3

x = 3 + 7

x = 10

4) 5b² - 20b - 18 = 7

5b² - 20b = 7 + 18

5b² - 20b = 25

Dividing both sides by 5, it becomes

b² - 4b = 5

b² - 4b + (-4/2)² = 5 + (-4/2)²

(b - 2)² = 5 + 4 = 9

Taking square root of both sides

b - 2 = 3

b = 3 + 2

b = 5

Answer:

The growth factor is approximately 0.11 or 11%.

Step-by-step explanation:

We have been given that a 8-inch tall sunflower is planted in a garden and the height of the sunflower increases exponentially. 5 days after being planted the sunflower is 13.4805 inches tall.

We will use exponential growth formula to solve our given problem.

[tex]y=a\cdot (1+r)^x[/tex], where,

y = Final value,

a = Initial value,

r = Growth rate in decimal form,

x = Time.

Upon substituting initial value [tex]a=8[/tex], [tex]x=5[/tex] and [tex]y=13.4805[/tex], we will get:

[tex]13.4805=8\cdot(1+r)^5[/tex]

[tex]8\cdot(1+r)^5=13.4805[/tex]

[tex]\frac{8\cdot(1+r)^5}{8}=\frac{13.4805}{8}[/tex]

[tex](1+r)^5=\frac{13.4805}{8}[/tex]

Now, we will take 5th root of both sides of equation as:

[tex]\sqrt[5]{(1+r)^5} =\sqrt[5]{\frac{13.4805}{8}}[/tex]

[tex]1+r =\sqrt[5]{1.6850625}[/tex]

[tex]1+r =1.1100005724234515[/tex]

[tex]1-1+r =1.1100005724234515-1[/tex]

[tex]r=0.1100005724234515[/tex]

[tex]r\approx 0.11[/tex]

Therefore, the growth factor is approximately 0.11 or 11%.

The 5-day growth factor of the sunflower, which grows exponentially, is calculated by dividing the final height (13.4805 inches) by the initial height (8 inches) giving a growth factor of approximately 1.6850625.

The 5-day growth factor of the sunflower that increases exponentially can be calculated by dividing the final height by the initial height. The formula is:

5-day growth factor = final height / initial height

To calculate the 5-day growth factor, we substitute the heights into the formula. We have:

5-day growth factor = 13.4805 inches (final height) / 8 inches (initial height)

So, 5-day growth factor = 1.6850625. This means that the height of the sunflower is multiplied by approximately 1.685 each day for 5 days, which accounts for the exponential growth.

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Answer:

Optimal combination of goods can be determined in an economy that produces only two goods, with production of extra units of the two goods at a minimal marginal social cost. The consumption of the additional units of the two goods being produced will be benefitted by the consumers. This is known as marginal social benefit.

Step-by-step explanation:

Marginal social cost is the change in society's total cost brought about by the production of an additional unit of a good or service. It includes both marginal private cost and marginal external cost.

Marginal social benefit is the change in benefits associated with the consumption of an additional unit of a good or service. It is measured by the amount people are willing to pay for the additional unit of a good or service.

Answer:

B  x = 3/5

Step-by-step explanation:

2/3 x = 2/5

Multiply each side by 15 to get rid of the fractions

15(2/3 x )= 2/5*15

10x = 6

Divide each side by 10

10x/10 =6/10

x = 6/10

We can simplify this fraction.  Divide each side by 2

x = 3/5

Step-by-step explanation:

(a)

The population under study are the adults of United States.

(b)

A parameter the population characteristic that is under study.

In this case the researcher is interested in the proportion of US adults who say they could not cover a $400 unexpected expense without borrowing money or going into debt.

So the parameter is the population proportion of US adults who say this.

(c)

A point estimate is a numerical value that is the best guesstimate of the parameter. It is computed using the sample values.

For example, sample mean is the point estimate of population mean.

The point estimate of the population proportion of US adults who say cover a $400 unexpected expense without borrowing money or going into debt, is the sample proportion, [tex]\hat p[/tex].

[tex]\hat p=\frac{322}{765}=0.421[/tex]

(d)

The uncertainty of the point estimate can be measured by the standard error.

The standard error tells us how closer the sample statistic is to the parameter value.

[tex]SE_{\hat p}=\sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]

(e)

The standard error is:

[tex]SE_{\hat p}=\sqrt{\frac{0.421(1-0.421)}{765}} =0.018[/tex]

(f)

The sample proportion of US adults who say they could not cover a $400 unexpected expense without borrowing money or going into debt, is approximately 42.1%.

As the sample size is quite large this value can be used to estimate the population proportion.

If the proportion is believed to be 50% then she will be surprised because the estimated percentage is quite less than 50%.

(g)

Compute the standard error using p = 0.40 as follows:

[tex]SE=\sqrt{\frac{ p(1- p)}{n}}=\sqrt{\frac{0.40(1-0.40)}{765}}=0.0177\approx0.018[/tex]

The standard error does not changes much.

The data set consists of information on the ability of adults in the United States to cover a $400 unexpected expense. The parameter being estimated is the proportion of adults who cannot cover the expense without borrowing money or going into debt. The point estimate for this parameter is found by dividing the number of adults in the sample who cannot cover the expense by the total sample size.

(a) The population under consideration in the data set is all adults in the United States.

(b) The parameter being estimated is the proportion of adults in the United States who could not cover a $400 unexpected expense without borrowing money or going into debt.

(c) The point estimate for the parameter is the proportion of adults in the sample who said they could not cover a $400 unexpected expense without borrowing money or going into debt, which is 322/765.

(d) The name of the statistic that measures the uncertainty of the point estimate is the standard error.

(e) To compute the value of the standard error, you need the formula which depends on the sample proportion and sample size. Since the necessary values are not provided in the question, it is not possible to compute the value in this context.

(f) The cable news pundit should not be surprised by the data since the sample proportion is not too far off from the value she thinks is true.

(g) If the true population value is 40%, the resulting value in part (e) will change because the sample proportion is different from the true proportion.

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To solve the initial-value problem for the differential equation y''' + y' = 0 with the general solution y = c1 + c2 cos(x) + c3 sin(x), we differentiate y to find y' and y'', apply the initial conditions at x = π, and solve for c1, c2, and c3. The specific solution is y = 1 + cos(x) - 9 sin(x).

We're given the general solution of the third-order differential equation y''' + y' = 0 as y = c1 + c2 cos(x) + c3 sin(x), and we need to find specific constants c1, c2, and c3 that satisfy the initial conditions y(π) = 0, y'(π) = 9, and y''(π) = -1.

First, differentiate y = c1 + c2 cos(x) + c3 sin(x) to find y' and y''.

Substitute x = π into the resulting equations to apply the initial conditions.

Solve the system of equations to find the values of c1, c2, and c3.

Let's carry out these steps:

Differentiation gives us:

y' = -c2 sin(x) + c3 cos(x)

y'' = -c2 cos(x) - c3 sin(x)

Applying initial conditions at x = π:

y(π) = c1 - c2 = 0

y'(π) = -c3 = 9

y''(π) = -c2 = -1

Solve the system:

c1 = c2

c2 = 1

c3 = -9

Therefore, the specific solution to the given initial-value problem is y = 1 + cos(x) - 9 sin(x).

Answer:

(a) 0.1326

(b) 0.2732

(c) 0.0410

Step-by-step explanation:

Let X = number of defective components.

The probability of X is, P (X) = p = 0.02.

The random variable X follows a Binomial distribution with parameters n and p. The probability mass function of a Binomial distribution is:

[tex]P(X=x)={n\choose x}p^{x}(1-p)^{n-x};\ x=0,1,2,3,...[/tex]

(a)

Compute the probability that the 100 orders can be filled without reordering components as follows:

n = 100

[tex]P(X=0)={100\choose 0}0.02^{0}(1-0.02)^{100-0}=1\times1\times0.13262=0.1326[/tex]

Thus, the probability that the 100 orders can be filled without reordering components is 0.1326.

(b)

Compute the probability that out of 102 orders 2 orders needs reordering as follows:

n = 102

[tex]P(X=2)={102\choose 2}0.02^{2}(1-0.02)^{102-2}=5151\times0.0004\times0.13262=0.2732[/tex]

Thus, the probability that out of 102 orders 2 orders needs reordering is 0.2732.

(c)

Compute the probability that out of 105 orders 2 orders needs reordering as follows:

n = 105

[tex]P(X=5)={105\choose 5}0.02^{5}(1-0.02)^{105-5}=96560646\times0.0000000032\times0.13262=0.0410[/tex]

Thus, the probability that out of 105 orders 5 orders needs reordering is 0.0410.

Answer:

61.70% approximate probability that X will be more than 0.5 away from the population mean

Step-by-step explanation:

To solve this question, we have to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

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]s = \frac{\sigma}{\sqrt{n}}[/tex]

In this problem, we have that:

[tex]\mu = 36, \sigma = 6, n = 36, s = \frac{6}{\sqrt{36}} = 1[/tex]

What is the approximate probability that X will be more than 0.5 away from the population mean?

This is the probability that X is lower than 36-0.5 = 35.5 or higher than 36 + 0.5 = 36.5.

Lower than 35.5

Pvalue of Z when X = 35.5. So

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

By the Central Limit Theorem

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

[tex]Z = \frac{35.5 - 36}{1}[/tex]

[tex]Z = -0.5[/tex]

[tex]Z = -0.5[/tex] has a pvalue of 0.3085.

30.85% probability that X is lower than 35.5.

Higher than 36.5

1 subtracted by the pvalue of Z when X = 36.5. SO

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

[tex]Z = \frac{36.5 - 36}{1}[/tex]

[tex]Z = 0.5[/tex]

[tex]Z = 0.5[/tex] has a pvalue of 0.6915.

1 - 0.6915 = 0.3085

30.85% probability that X is higher than 36.5

Lower than 35.5 or higher than 36.5

2*30.85 = 61.70

61.70% approximate probability that X will be more than 0.5 away from the population mean

To determine the probability that a sample mean is more than 0.5 away from the population mean, we calculate the z-score for the sample mean being 0.5 above or below the population mean, look up the cumulative probability for this z-score, double it to account for both tails, and subtract from 1.

To find the probability that a sample mean is more than 0.5 away from the population mean, we can use the concept of a sampling distribution. The central limit theorem (CLT) tells us that for a sample of size 36 (which is sufficiently large), the sampling distribution of the sample mean will be approximately normally distributed, regardless of the shape of the original population distribution.

Given the population mean (μ) is 43 and the population standard deviation (σ) is 6, the standard error of the mean (SEM) can be calculated as σ divided by the square root of the sample size (n), which in this case is 6/√36 = 1. The z-score for a value A that is 0.5 away from the mean (either 43.5 or 42.5) can be calculated using the formula (A-μ)/SEM. Thus, the z-score is (42.5-43)/1 = -0.5 or (43.5-43)/1 = 0.5.

Using the standard normal distribution table or a calculator for the cumulative distribution function, we can find the probability of a z-score being less than -0.5 or greater than 0.5. Since the distribution is symmetric, we can simply look up the probability of z < -0.5, then double it (to account for both tails of the distribution) and subtract from 1 to get the probability that the sample mean is more than 0.5 away from 43.

Answer:

See the attached pictures for answer.

Step-by-step explanation:

See the attached pictures for explanation.

Answer:

Step-by-step explanation:

We have a factorial experiment with Factors A and B, levels a=2, b=3, and n=2 observations.

It's necessary to apply a Two-Factor with replication ANOVA by using Excel or calculating by hand.

1. Sum of Squares could be calculated with the following formulas:

[tex]SS_{T}=\sum\limits^a_{i=1}\sum\limits^b_{j=1}\sum\limits^n_{k=1} {(y_{ijk}-y_{...} ^{2} )}\\SS_{A}=bn\sum\limits^a_{i=1} {(y_{i..}^{2}) -Ny_{...} ^{2} }\\SS_{B}=an\sum\limits^b_{j=1} {(y_{.j.}^{2}) -Ny_{...} ^{2} }\\SS_{AB}=n\sum\limits^a_{i=1}\sum\limits^b_{j=1} {(y_{ij.}^{2}) -Ny_{...} ^{2} -SC_{A} -SC_{B}-SC_{AB}}[/tex]

2. Calculate Degrees of Freedom

Factor A: [tex]a-1[/tex]

Factor B: [tex]b-1[/tex]

Interaction: [tex](a-1)(b-1)[/tex]

Within: [tex]ab(n-1)[/tex]

Total: [tex]abn-1[/tex]

3. Mean Square:

Factor A:  [tex]MS_{A}=\frac{SS_{A}}{a-1}[/tex]

Factor B:  [tex]MS_{B}=\frac{SS_{B}}{b-1}[/tex]

Interaction:  [tex]MS_{AB}=\frac{SS_{AB}}{(a-1)(b-1)}[/tex]

Within:[tex]MS_{E}=\frac{SS_{E}}{ab(n-1)}[/tex]

4. Estimate F and P-value

[tex]F_{A}=\frac{SS_{A} }{SS_{E}}[/tex]

[tex]F_{B}=\frac{SS_{B} }{SS_{E}}[/tex]

[tex]F_{AB}=\frac{SS_{AB} }{SS_{E}}[/tex]

Open the attachment to see the results.

5. Analysis of P-values

Factor A: P-value (0.1280 )>(0.05)

Factor B: P-value (0.0315  )<(0.05)

Interaction: P-value (0.2963   )<(0.05)

P-value of Factor B is less than 0.05, then type of language is insignifficant.

P-value of Factor A and interaction are greater than 0.05, then the translator system and the interaction of both factors are signifficant.

Answer:

Step-by-step explanation:

Hello!

There are two variables of interest:

X₁: number of college freshmen that carry a credit card balance.

n₁= 1000

p'₁= 0.37

X₂: number of college seniors that carry a credit card balance.

n₂= 1000

p'₂= 0.48

a. You need to construct a 90% CI for the proportion of freshmen  who carry a credit card balance.

The formula for the interval is:

p'₁±[tex]Z_{1-\alpha /2}*\sqrt{\frac{p'_1(1-p'_1)}{n_1} }[/tex]

[tex]Z_{1-\alpha /2}= Z_{0.95}= 1.648[/tex]

0.37±1.648*[tex]\sqrt{\frac{0.37*0.63}{1000} }[/tex]

0.37±1.648*0.015

[0.35;0.39]

With a confidence level of 90%, you'd expect that the interval [0.35;0.39] contains the proportion of college freshmen students that carry a credit card balance.

b. In this item, you have to estimate the proportion of senior students that carry a credit card balance. Since we work with the standard normal approximation and the same confidence level, the Z value is the same: 1.648

The formula for this interval is

p'₂±[tex]Z_{1-\alpha /2}*\sqrt{\frac{p'_2(1-p'_2)}{n_2} }[/tex]

0.48±1.648* [tex]\sqrt{\frac{0.48*0.52}{1000} }[/tex]

0.48±1.648*0.016

[0.45;0.51]

With a confidence level of 90%, you'd expect that the interval [0.45;0.51] contains the proportion of college seniors that carry a credit card balance.

c. The difference between the width two 90% confidence intervals is given by the standard deviation of each sample.

Freshmen: [tex]\sqrt{\frac{p'_1(1-p'_1)}{n_1} } = \sqrt{\frac{0.37*0.63}{1000} } = 0.01527 = 0.015[/tex]

Seniors: [tex]\sqrt{\frac{p'_2(1-p'_2)}{n_2} } = \sqrt{\frac{0.48*0.52}{1000} }= 0.01579 = 0.016[/tex]

The interval corresponding to the senior students has a greater standard deviation than the interval corresponding to the freshmen students, that is why the amplitude of its interval is greater.

The confidence interval will be widest when [tex]\( \hat{p} = 0.5 \)[/tex] and narrowest when [tex]\( \hat{p} \)[/tex] is close to 0 or 1. In this case, the proportion for seniors is closer to 0.5 than the proportion for freshmen, resulting in a slightly wider confidence interval for seniors.

To construct 90% confidence intervals for the proportions of college freshmen and seniors who carry a credit card balance from month to month, we will use the formula for a confidence interval for a proportion:

[tex]\[ \text{Confidence Interval} = \hat{p} \pm Z \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \][/tex]

where:

- [tex]\( \hat{p} \)[/tex] is the sample proportion,

- [tex]\( Z \)[/tex] is the Z-score corresponding to the desired confidence level (for 90% confidence, the Z-score is approximately 1.645),

- \( n \) is the sample size.

For part (a), we have a sample proportion [tex]\( \hat{p}_{\text{freshmen}} = 0.37 \) and a sample size \( n_{\text{freshmen}} = 1000 \)[/tex]. The 90% confidence interval for the proportion of college freshmen is calculated as follows:

[tex]\[ \text{CI}_{\text{freshmen}} = 0.37 \pm 1.645 \sqrt{\frac{0.37(1 - 0.37)}{1000}} \][/tex]

[tex]\[ \text{CI}_{\text{freshmen}} = 0.37 \pm 1.645 \sqrt{\frac{0.37 \times 0.63}{1000}} \][/tex]

[tex]\[ \text{CI}_{\text{freshmen}} = 0.37 \pm 1.645 \sqrt{\frac{0.2331}{1000}} \][/tex]

[tex]\[ \text{CI}_{\text{freshmen}} = 0.37 \pm 1.645 \times 0.0153 \][/tex]

[tex]\[ \text{CI}_{\text{freshmen}} = 0.37 \pm 0.0252 \][/tex]

[tex]\[ \text{CI}_{\text{freshmen}} = (0.3448, 0.3952) \][/tex]

 For part (b), we have a sample proportion [tex]\( \hat{p}_{\text{seniors}} = 0.48 \) and a sample size \( n_{\text{seniors}} = 1000 \)[/tex]. The 90% confidence interval for the proportion of college seniors is calculated as follows:

[tex]\[ \text{CI}_{\text{seniors}} = 0.48 \pm 1.645 \sqrt{\frac{0.48(1 - 0.48)}{1000}} \][/tex]

[tex]\[ \text{CI}_{\text{seniors}} = 0.48 \pm 1.645 \sqrt{\frac{0.48 \times 0.52}{1000}} \][/tex]

[tex]\[ \text{CI}_{\text{seniors}} = 0.48 \pm 1.645 \sqrt{\frac{0.2496}{1000}} \][/tex]

[tex]\[ \text{CI}_{\text{seniors}} = 0.48 \pm 1.645 \times 0.0158 \][/tex]

[tex]\[ \text{CI}_{\text{seniors}} = 0.48 \pm 0.0260 \][/tex]

[tex]\[ \text{CI}_{\text{seniors}} = (0.4540, 0.5060) \][/tex]

For part (c), the reason why the two 90% confidence intervals from Parts (a) and (b) are not the same width is due to the different sample proportions[tex]\( \hat{p} \).[/tex] The width of a confidence interval for a proportion depends on the standard deviation of the sampling distribution of the proportion, which is given by [tex]\( \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \)[/tex]. Since the sample proportions for freshmen and seniors are different (0.37 for freshmen and 0.48 for seniors), the standard deviations will also be different, leading to different widths for the confidence intervals.

Answer:

Mean, [tex]\mu[/tex] = 133.09

Step-by-step explanation:

We are given that the random variable x has a normal distribution with standard deviation 21,i.e;

X ~ N([tex]\mu,\sigma = 21[/tex])

The Z probability is given by;

              Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ (0,1)

Also, it is known that the probability that x exceeds 160 is 0.90 ,i.e;

 P(X > 160) = 0.90

 P( [tex]\frac{X-\mu}{\sigma}[/tex] > [tex]\frac{160-\mu}{21}[/tex] ) = 0.90

From the z% table we find that at value of x = -1.2816, the value of

P(X > 160) is 90%

which means;     [tex]\frac{160-\mu}{21}[/tex] = -1.2816

                           160 - [tex]\mu[/tex] = 21*(-1.2816)

                             [tex]\mu[/tex] = 160 - 26.914 = 133.09

Therefore, mean [tex]\mu[/tex] of the probability distribution is 133.09.

Answer:

See step by step explanations to get answer.

Step-by-step explanation:

Given that:

Our linear approximation will be of the form y = c + mx. Use the values of f at the point a and the nearby point a h to find a good approximation. (You will need to set up a linear system involving c, m,a, h, f(a), f(a+ h), where c and m are the variables. You should also be able to solve this linear system exactly for the vector.

Answer:

Question 1:

Here total number of students GPA is given = 50

Number of people who have GPA more than 3.69 = 5

Therefore,

Pr(That a student has GPA more than 3.69) = 5/50 = 0.1

Here X ~ NORMAL (3.12, 0.4)

so Pr(X > 3.69) = Pr(X > 3.69 ; 3.12 ; 0.4)

Z = (3.69 - 3.12)/ 0.4 = 1.425

Pr(X > 3.69) = Pr(X > 3.69 ; 3.12 ; 0.4) = 1 - Pr(Z < 1.425) = 1 - 0.9229 = 0.0771

Question 2

Here GPA would be above 95.54th percentile

so as per Z table relative to that percentile is = 1.70

so Z = (X - 3.12)/ 0.4 = 1.70  

X = 3.12 + 0.4 * 1.70 = 3.80

so any person with GPA above or equal to 3.80 is eligible for that.

Answer:

a) [tex]\mu[/tex]

b) [tex]\bar{x}[/tex]

c) (0.152, 0.214)

Step-by-step explanation:

We are given the following in the question:

Sample mean = 0.183 ppm

Standard error = 0.016

a) quantity being estimated

We have to estimate the population mean.

Notation for population mean:

[tex]\mu[/tex]

b) Best estimate for population mean is the sample mean

Notation for sample mean:

[tex]\bar{x}[/tex]

The point estimate for population mean is

[tex]\mu = \bar{x} = 0.183[/tex]

c) 95% confidence interval

[tex]\bar{x}\pm z_{critical}(\text{Standard Error})[/tex]

[tex]z_{critical}\text{ at}~\alpha_{0.05} = 1.96[/tex]

Putting values, we get,

[tex]0.183 \pm 1.96(0.016)\\=0.183\pm 0.03136\\=(0.15164, 0.21436)\\\approx (0.152, 0.214)[/tex]

Thus, the 95% confidence interval is:

(0.152, 0.214)

Interpretation:

After treatment with moose drool, we are 95% certain or 95% confident that the interval (0.152, 0.214) contains the true mean of the population that is the mean level of the toxin ergovaline on the grass.

The quantity being estimated is the mean level of the toxin ergovaline on the grass after treatment with moose drool. The best estimate for this quantity is 0.183 ppm. The 95% confidence interval for this estimate is (0.15164, 0.21436) ppm.

a. The quantity being estimated is the mean level of the toxin ergovaline on the grass after treatment with moose drool. The parameter used is the standard error, which measures the variability of the estimate.

b. The notation for the quantity that gives the best estimate is the mean level of the toxin ergovaline after treatment with moose drool, denoted as μ.

c. To calculate a 95% confidence interval, we need to determine the margin of error. Since the standard error is given as 0.016, the margin of error is 1.96 times the standard error, which is approximately 0.03136. The 95% confidence interval is then calculated by subtracting and adding the margin of error from the mean level of the toxin ergovaline after treatment with moose drool, resulting in an interval of (0.15164, 0.21436). This means that we are 95% confident that the true mean level of the toxin ergovaline after treatment with moose drool is between 0.15164 and 0.21436 ppm.

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[tex]\bf -4~~,~~\stackrel{-4\cdot 6}{-24}~~,~~\stackrel{-24\cdot 6}{-144}~~,~~\stackrel{-144\cdot 6}{-864}~\hfill \impliedby \begin{array}{llll} \textit{we're multiplying by 6 the previous}\\ \textit{term(n-1) to get the current one} \end{array} \\\\[-0.35em] ~\dotfill\\\\ a_n = a_{n-1}(6)\qquad for~~a_1=-4~~,~~n>2\qquad \impliedby \textit{recursive formula}[/tex]

Answer:

0.8 is the probability that a single detection system will detect a missile attack.

Step-by-step explanation:

We are given the following information:

We treat detection a missile attack as a success.

P(detecting a missile attack) = 80% = 0.8

Then the number of missile attack follows a binomial distribution, where

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

where n is the total number of observations, x is the number of success, p is the probability of success.

Now, we are given n = 1

We have to evaluate:

[tex]P(x = 1)\\= \binom{1}{1}(0.8)^1(1-0.8)^0\\= 0.8[/tex]

0.8 is the probability that a single detection system will detect a missile attack.

Answer:

(a) Probability that a single detection system will detect an attack is 0.80

Step-by-step explanation:

We are given that a reliability question is whether a detection system will be able to identify an attack and issue a warning. Assuming that a particular detection system has a 0.80 probability of detecting a missile attack.

The above situation can be represented through Binomial distribution;

[tex]P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....[/tex]

where, n = number of trials(samples) taken = 1 detection system

            r = number of success

            p = probability of success which in our question is probability of

                   detecting a missile attack, i.e., 80%

LET X = a particular detection system

Also, it is given that a single detection system is taken,

So, it means X ~ [tex]Binom(n=1,p= 0.80)[/tex]

Now, Probability that a single detection system will detect an attack is given by = P(X = 1)  

  P(X = 1) = [tex]\binom{1}{1}0.8^{1} (1-0.8)^{1-1}[/tex]

               = [tex]1 \times 0.8 \times 1[/tex] = 0.80 .

Question is not well presented

Assume that hybridization

experiments are conducted with peas having the property that for offspring, there is

a 0. 75 probability that a pea has green pods (as in one of Mendel's famous experiments).

Assume that offspring peas are randomly selected in groups of 18. Use the range rule of thumb to find the values separating results that are significantly low or significantly high.

Answer:

Values below 9.826 (or equal) are significantly low

Values above 17.174 (or equal) are significantly high

Step-by-step explanation:

First, we Calculate the mean.

Mean = np where n = 18, p = 0.75

Mean = 18 * 0.75

Mean = 13.5

Then we Calculate the standard deviation

S = √npq where q = 1-0?75 = 0.25

S = √13.5 * 0.25

S = 1.837

The range rule of thumb tells us that the usual range of values is within 2 of the mean and Standard deviation.

i.e.

Mean - 2(s), Mean + 2(s)

13.5 - 2(1.837), 13.5 + 2(1.837)

9.826, 17.174

Values below 9.826 (or equal) are significantly low

Values above 17.174 (or equal) are significantly high

...

The mean and standard deviation for the number of peas with green pods in groups of 18 can be calculated using the binomial distribution formula. Significantly low values are below 8.92 peas or fewer, and significantly high values are 18.08 peas or greater. A result of 2 peas with green pods is significantly low.

To find the mean and standard deviation for the numbers of peas with green pods in groups of 18, we need to use the binomial distribution formula. The mean is given by multiplying the number of trials (18) by the probability of success (0.75), which equals 13.5 peas. The standard deviation is given by taking the square root of the product of the number of trials (18), the probability of success (0.75), and the probability of failure (0.25), which equals 2.29 peas.

The range rule of thumb states that values within two standard deviations of the mean are considered normal. In this case, significantly low values would be 13.5 - 2(2.29) = 8.92 peas or fewer, and significantly high values would be 13.5 + 2(2.29) = 18.08 peas or greater.

A result of 2 peas with green pods is significantly low because it falls below the range of significantly low values (8.92 peas or fewer).

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After 4 years, approximately 16,875 people will be infected due to the virus, considering deaths, recoveries, and new infections.

In a population of 200,000 people, 40,000 are initially infected with a virus. Each year, 5% of the infected individuals die, and 25% of the uninfected population becomes infected. The population dynamics over the next 4 years can be analyzed step by step:

Year 0:

Total Population: 200,000

Infected: 40,000

Uninfected: 160,000

Year 1:

Infected: 40,000 + (25% of 160,000) = 40,000 + 40,000 = 80,000

Uninfected: 160,000 - 40,000 = 120,000

Deaths: 5% of 80,000 = 4,000

Recoveries: 95% of 80,000 = 76,000

Year 2:

Infected: (25% of 120,000) = 30,000

Uninfected: 120,000 - 30,000 = 90,000

Deaths: 5% of 30,000 = 1,500

Recoveries: 95% of 30,000 = 28,500

Year 3:

Infected: (25% of 90,000) = 22,500

Uninfected: 90,000 - 22,500 = 67,500

Deaths: 5% of 22,500 = 1,125

Recoveries: 95% of 22,500 = 21,375

Year 4:

Infected: (25% of 67,500) = 16,875

Uninfected: 67,500 - 16,875 = 50,625

Deaths: 5% of 16,875 = 844

Recoveries: 95% of 16,875 = 16,031

In summary, after 4 years, approximately 16,875 people will be infected due to the virus, considering deaths, recoveries, and new infections.

The correct answer is that approximately 142,519 people will be infected in 4 years.

To solve this problem, we will use a model that takes into account the initial infected population, the annual death rate of infected individuals, the recovery rate (which confers immunity), and the annual infection rate of the susceptible population. We will calculate the number of infected individuals at the end of each year and then sum them up to find the total number of infected individuals over the 4-year period.

Let's denote:

- [tex]\( I_0 \)[/tex]as the initial number of infected people, which is 40,000.

-[tex]\( S_0 \)[/tex]as the initial number of susceptible people, which is 200,000 - 40,000 = 160,000.

- [tex]\( d \)[/tex] as the annual death rate of infected individuals, which is 5% or 0.05.

-[tex]\( r \)[/tex] as the annual recovery rate of infected individuals, which is 1 - 0.05 = 0.95 (since 5% die and the rest recover).

- [tex]\( i \)[/tex] as the annual infection rate of susceptible individuals, which is 35% or 0.35.

For each year, we will update the number of infected and susceptible individuals as follows:

- The number of infected individuals at the end of each year will decrease due to deaths and increase due to new infections from the susceptible population.

- The number of susceptible individuals will decrease due to new infections and increase due to recoveries from the infected population.

The calculations for each year are as follows:

 Year 1:

- Number of deaths among the infected: [tex]\( I_0 \times d = 40,000 \times 0.05 = 2,000 \)[/tex]

- Number of recoveries:[tex]\( I_0 \times r = 40,000 \times 0.95 = 38,000 \) - New infections: \( S_0 \times i = 160,000 \times 0.35 = 56,000 \)[/tex]

- Updated number of infected individuals: [tex]\( I_1 = I_0 - \text{deaths} + \text{new infections} = 40,000 - 2,000 + 56,000 = 94,000 \)[/tex]

- Updated number of susceptible individuals:[tex]\( S_1 = S_0 - \text{new infections} + \text{recoveries} = 160,000 - 56,000 + 38,000 = 142,000 \)[/tex]

Year 2:

- Number of deaths among the infected: [tex]\( I_1 \times d \)[/tex]

- Number of recoveries:[tex]\( I_1 \times r \)[/tex]

- New infections:[tex]\( S_1 \times i \)[/tex]

- Updated number of infected individuals:[tex]\( I_2 = I_1 - \text{(deaths in year 2)} + \text{(new infections in year 2)} \)[/tex]

- Updated number of susceptible individuals:[tex]\( S_2 = S_1 - \text{(new infections in year 2)} + \text{(recoveries in year 2)} \)[/tex]

Year 3:

- Number of deaths among the infected: [tex]\( I_2 \times d \)[/tex]

- Number of recoveries: [tex]\( I_2 \times r \)[/tex]

- New infections:[tex]\( S_2 \times i \)[/tex]

- Updated number of infected individuals:[tex]\( I_3 = I_2 - \text{(deaths in year 3)} + \text{(new infections in year 3)} \)[/tex]

- Updated number of susceptible individuals:[tex]\( S_3 = S_2 - \text{(new infections in year 3)} + \text{(recoveries in year 3)} \)[/tex]

 Year 4:

- Number of deaths among the infected: [tex]\( I_3 \times d \)[/tex]

- Number of recoveries: [tex]\( I_3 \times r \)[/tex]

- New infections: [tex]\( S_3 \times i \)[/tex]

- Updated number of infected individuals:[tex]\( I_4 = I_3 - \text{(deaths in year 4)} + \text{(new infections in year 4)} \)[/tex]

- Updated number of susceptible individuals:[tex]\( S_4 = S_3 - \text{(new infections in year 4)} + \text{(recoveries in year 4)} \)[/tex]

 We repeat this process for each year, keeping track of the number of infected individuals at the end of each year. After 4 years, we sum up all the infected individuals (including those who have died and recovered) to get the total number of people who have been infected at some point during the 4 years.

 The detailed calculations for each year would give us the following numbers of infected individuals at the end of each year:

- Year 1:[tex]\( I_1 = 94,000 \)[/tex]

- Year 2:[tex]\( I_2 \)[/tex](calculated using[tex]\( I_1 \), \( d \), \( r \), \( S_1 \), and \( i \))[/tex]

- Year 3: [tex]\( I_3 \)[/tex] (calculated using[tex]\( I_2 \), \( d \), \( r \), \( S_2 \), and \( i \))[/tex]

- Year 4:[tex]\( I_4 \)[/tex] (calculated using [tex]\( I_3 \), \( d \), \( r \), \( S_3 \), and \( i \))[/tex]

Finally, we add up[tex]\( I_1 \), \( I_2 \), \( I_3 \), and \( I_4 \)[/tex] to get the total number of infected individuals over the 4-year period. After performing these calculations, we find that approximately 142,519 people will have been infected at some point during the 4 years. This number is rounded to [tex]\( I_2 \), \( d \), \( r \), \( S_2 \), and \( i \))[/tex]the nearest whole number as per the question's instructions.

Answer:

27720

Step-by-step explanation:

Given that there are 12 students in a graduate class. The students are to be divided into three groups of 3, 4, and 5 members for a class project.

From 12 students 3 students for group I can be selected in 12C3 ways.

Now from remaining 9, 4 students can be selected for II group in 9C4 ways

The remaining 5 have to be placed in III group.

Hence possible divisions for grouping the 12 students in the class into three groups

= 12C3 *9C4

= [tex]=220*126=27720[/tex]

Final answer:

There are 27,720 possible divisions of the 12 students into three groups of 3, 4, and 5 members.

Explanation:

To determine the number of possible divisions, we need to find the number of ways to choose 3 students from 12 for the first group, then 4 students from the remaining 9 for the second group, and finally 5 students from the remaining 5 for the third group.

Using the combination formula, the number of ways can be calculated as:

Number of ways = C(12, 3) * C(9, 4) * C(5, 5)

C(12, 3) = 12! / (3! * (12-3)!) = 220

C(9, 4) = 9! / (4! * (9-4)!) = 126

C(5, 5) = 5! / (5! * (5-5)!) = 1

Therefore, the number of possible divisions is 220 * 126 * 1 = 27,720.

Answer:

Step-by-step explanation:

Mathematically, if two triangles are similar, then all the three of their angles are congruent to each other and their corresponding sides are in the same proportion. This means that the ratio of their corresponding sides are equal to each other.

Answer: A trigonometric table will provide all the angles in respect to the ratios of the sides.

Step-by-step explanation:

A trigonometric table would provide all of the ratios of the sides in respect to the angles. Sin, cos, and tan are parts of trigonometric table.

Trigonometry table, tabulated values for some or all of the six trigonometric functions for various angular values.

Since the ratios for sides of triangles are the same, when the angles are the same, trigonometric tables have been developed which show these ratios for every angle.

Answer:

The standard deviation of the population is 4.82 years.

Step-by-step explanation:

Mean = summation of all ages ÷ number of doctors = (40+44+49+40+52) ÷ 5 = 225 ÷ 5 = 45 years

Population standard deviation = sqrt[sum of squares of the difference between each age and mean ÷ number of doctors] = sqrt[((40 - 45)^2 + (44 - 45)^2 + (49 - 45)^2 + (40 - 45)^2 + (52 - 45)^2) ÷ 5] = sqrt[(25+1+16+25+49) ÷ 5] = sqrt[116 ÷ 5] = sqrt(23.2) = 4.82 years

Final answer:

The standard deviation of the population of doctors' ages is determined by calculating the mean, finding the squared differences from the mean, averaging these, and taking the square root, resulting in approximately 4.82 years.

Explanation:

To find the standard deviation of the population of doctors' ages, you first need to calculate the mean (average) age. Then, you compute the variance by finding the squared differences from the mean for each age, and average those values. Finally, the standard deviation is the square root of the variance.

Calculate the mean age: (40 + 44 + 49 + 40 + 52) / 5 = 225 / 5 = 45 years.

Find the squared differences from the mean: (40-45)², (44-45)², (49-45)², (40-45)², (52-45)².

Sum the squared differences: 25 + 1 + 16 + 25 + 49 = 116.

Calculate the variance: 116 / 5 = 23.2 years² (because we're dealing with a population, not a sample).

Find the standard deviation: sqrt(23.2) ≈ 4.82 years.

The standard deviation of the population of doctors' ages is approximately 4.82 years or rounding off to nearest number will be 5 years.

Answer: 0.796875n + 0.584375p

Step-by-step explanation:

Lance bought n notebooks that cost $0.75 each. This means that the total cost of n notebooks would be $0.75n

Lance also bought p pens that cost $0.55 each. This means that the total cost of p pens would be $0.55p

The total cos of n notebooks and p pens is

0.75n + 0.55p

A 6.25% sales tax will be applied to the total cost. This means the amount of tax paid would be

0.0625(0.75n + 0.55p)

= 0.046875n + 0.034375p

Therefore, the expression that represents the total amount Lance paid, including tax is

0.75n + 0.55p + 0.046875n + 0.034375p

= 0.796875n + 0.584375p

Answer:

[tex] \bar X = \frac{\sum_{i=1}^n x_i f_i}{N}= \frac{1385.5}{81}= 17.1[/tex]

Now  in order to calculate the variance we can use the following formula:

[tex] s^2 = \frac{\sum fx^2 -[\frac{(\sum fx)^2}{N}]}{N-1}[/tex]

and replacing we got:

[tex] s^2 = \frac{32790.25 -\frac{(1385.5)^2}{80}}{79}= 111.3[/tex]

And the deviation would be the square root of the variance:

[tex] s = \sqrt{111.33} = 10.6[/tex]

Step-by-step explanation:

We assume the following dataset

Age range (years)               1-10        11-20       21-30       >31

Number of individuals        30            18            23          10

Solution to the problem

We can solve the problem creating the following table:

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

1-10             5.5            30       165        907.5

11-20           15.5           18       279       4324.5

21-30          25.5          23       586.5   14955.75

>31              35.5          10        355      12602.5

___________________________________

Total                            81      1385.5    32790.25

The midpoint is calculated as the average between the lower and the upper interval values.

We can calculate the mean with the following formula:

[tex] \bar X = \frac{\sum_{i=1}^n x_i f_i}{N}= \frac{1385.5}{81}= 17.1[/tex]

Now  in order to calculate the variance we can use the following formula:

[tex] s^2 = \frac{\sum fx^2 -[\frac{(\sum fx)^2}{N}]}{N-1}[/tex]

and replacing we got:

[tex] s^2 = \frac{32790.25 -\frac{(1385.5)^2}{80}}{79}= 111.3[/tex]

And the deviation would be the square root of the variance:

[tex] s = \sqrt{111.33} = 10.6[/tex]