We have a biased coin (probability of heads is equal to 1/4). Consider the following 2 step process: In the first step we flip the coin until we get a heads. Let X denote the trial on which the first heads occurs. In the second step we flip the coin X more times. Let Y be the number of heads in the second step. (a) For each non-negative integer, k, what is the probability that X = k? (b) Conditioned on the event {X = k}. What is the probability Y = 0? (c) Use the Law of Total Probability to compute the unconditional probability that Y = 0.

The probability that the market research predicts an unsuccessful market for the product and the product is actually unsuccessful is 24 percent. This is calculated using the concept of conditional probability and multiplying the probability of the product actually failing (30%) with the probability of market research correctly predicting a failure (80%).

To calculate the probability that the market research indicates an unsuccessful market for the product and the product is actually unsuccessful, we first need to understand the concept of conditional probability, given certain conditions.

Here, there is a 30 percent chance that a product will fail. The market research incorrectly predicts success for failed products 20 percent of the time. Therefore, the market research predicts failure for failed products 80 percent of the time (100% - 20%).

By multiplying these probabilities together, we can find the joint probability that a product is unsuccessful and market research also predicts it to be unsuccessful.

The result is 0.30 x 0.80 = 0.24. Therefore, there is a 24 percent chance that the market research will predict a failure and the product will also actually be a failure.

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The probability that market research indicates an unsuccessful market and the product is unsuccessful is 24%.

To calculate the probability that the results indicate an unsuccessful market for the product and that the product is unsuccessful, we need to use Bayes' Theorem. Initially, we have the probability of a product being successful, P(Success) = 0.70, and the probability of a product being a failure, P(Failure) = 0.30. Furthermore, if the product is successful, market research predicts it successful 90% of the time. If the product is a failure, market research incorrectly predicts it as a success 20% of the time, meaning it correctly predicts failure 80% of the time (100% - 20%).

We are interested in P(Market Research Failure and Actual Failure). This is the probability that the market research indicates a failure and the product indeed fails. Using the probabilities provided:

Therefore, P(Market Research Failure and Actual Failure) can be calculated as:

P(Market Research Failure and Actual Failure) = P(Market Research Failure | Failure) x P(Failure) = 0.80 x 0.30 = 0.24 or 24%.

Answer:

Step-by-step explanation:

We would apply the formula for determining compound interest which is expressed as

y = P(1 + r/n)^nt

Where

y = the value of the investment at the end of t years

r represents the interest rate.

n represents the periodic interval at which it was compounded.

P represents the principal or initial amount invested

From the information given,

P = $4700

r = 4.75% = 4.75/100 = 0.0475

n = 1 because it was compounded once in a year.

Therefore, the exponential function showing the relationship between y and t is

y = 4700(1 + 0.0475/1)^1 × t

y = 4700(1.0475)^t

The function f(x) = x3 - 9x2 - 21x + 8 increases in the intervals (-∞, -1) and (7, ∞), and decreases in the interval (-1, 7). It achieves a local minimum at -190 and a local maximum at -23. The inflection point is at x = 3, with the function being concave up for x < 3 and concave down for x > 3.

Given the function f(x) = x3 - 9x2 - 21x + 8, we need to find intervals where the function is increasing or decreasing, and determine its local minima/maxima, inflection points, and where it is concave up or down.

To find the intervals, we need to first take the derivative of f(x): f'(x) = 3x2 - 18x - 21 which becomes 3(x - 7)(x + 1). The critical points are then 7 and -1.

The function increases in the interval (-∞, -1) and decreases in the interval (-1, 7). The function again increases in the interval (7, ∞).

Local minima and maxima can be found by evaluating the function at the critical points. f(-1) = -23 and f(7) = -190, so the local maximum value of f is -23 and local minimum is -190.

To find the inflection point, we compute the second derivative f''(x) = 6x - 18, which simplifies to 6(x - 3), giving us an inflection point at x = 3.

The function is concave up for x < 3 and concave down for x > 3.

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a) f(x) is increasing for x in (- ∞, - 1) cup(7, infty)

f(x) is decreasing for x \in (- 1, 7)

b)Local minimum=-237 Local maximum=19

c)Point of inflection: (x, y) = (3, - 109)

d)Interval for concave upward curve: x \in (3, ∞)

e)Interval for concave upward curve: x \in (- ∞, 3)

The given function is [tex]f(x) = x ^ 3 - 9x ^ 2 - 21x + 8[/tex]

[tex]f' * (x) = 3x ^ 2 - 18x - 21[/tex]

f^ prime prime (x) = 6x - 18

a)f(x) is increasing for

f' * (x) > 0

[tex]3x ^ 2 - 18x - 21 > 0[/tex]

[tex]x ^ 2 - 6x - 7 > 0[/tex]

[tex]x ^ 2 - 7x + x - 7 > 0[/tex]

x(x - 7) + 1(x - 7) > 0

(x - 7)(x + 1) > 0

x > 7 and x > - 1or x < 7 and x < - 1

f(x) is decreasing for

f' * (x) < 0

3x ^ 2 - 18x - 21 < 0

x ^ 2 - 6x - 7 < 0

x ^ 2 - 7x + x - 7 < 0

x(x - 7) + 1(x - 7) < 0

(x - 7)(x + 1) < 0

x > 7 and x < - 1 or x < 7 and x > - 1

x=(-1,7)

b)for local extrema

f' * (x) = 0

3x ^ 2 - 18x - 21 = 0

x ^ 2 - 6x - 7 < 0

x ^ 2 - 7x + x - 7 = 0

x(x - 7) + 1(x - 7) = 0

(x - 7)(x + 1) = 0

x = 7 and x = - 1

At x=-1:

f(- 1) = (- 1) ^ 3 - 9 * (- 1) ^ 2 - 21(- 1) + 8 = 19

At x=7:

f(7) = (7) ^ 3 - 9 * (7) ^ 2 - 21(7) + 8 = - 237

Local minimum=-237

Local maximum=19

c) For point of inflection:

f^ prime prime (x) = 0

Rightarrow 6x - 18 = 0

Rightarrow x = 3

y = f(3) = (3) ^ 3 - 9 * (3) ^ 2 - 21(3) + 8 = - 109

(x, y) = (3, - 109)

d)Interval for concave upward curve:

f^ prime prime (x) > 0

6x - 18 > 0

x \in (3, ∞)

e)Interval for concave upward curve:

f^ prime prime (x) < 0

6x - 18 < 0

X€(−0,3)

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Answer:

3/20

Step-by-step explanation:

[tex]\frac{3}{5}[/tex]  ÷  4    =     [tex]\frac{3}{5}[/tex] [tex]* \frac{1}{4}[/tex]    =      [tex]\frac{3 * 1}{5 * 4}[/tex]     =     [tex]\frac{3}{20}[/tex]

To calculate the fraction of the class that earned an A, multiply 3/5 by 1/4, which equals 3/20. Thus, 3/20 of the class earned an A. This is the correct answer among the given options.

Calculating the Fraction of a Class That Earned an A

To find out what fraction of the class earned an A, we start with the fact that 3/5 of the class earned an A or a B.

We also know that 1/4 of those students earned an A.

To determine the fraction of the entire class that earned an A, we multiply these two fractions together:

Fraction of class that earned an A = (3/5) × (1/4)

This multiplication gives us:

(3 × 1) / (5 × 4) = 3/20

Therefore, 3/20 of the class earned an A.

This fraction cannot be reduced further and it is also one of the options provided, confirming that it is the correct answer.

Answer:

[tex]a. \ P(X\leq 8)=0.932\\b. \ P(X=8)=0.065\\c. \ P(9\leq X)=0.068\\d. \ P(5\leq X \leq 8)=0.491\\[/tex]

e. P(5<X<8)=0.251

Step-by-step explanation:

Given that boiler follows a Poisson distribution,$\sim$

[tex]X $\sim$Poi(\mu=5)[/tex]

Poisson Distribution formula is given by the expression:-

[tex]p(x,\mu)=\frac{e^{-\mu}\mu^x}{x!}[/tex]

Our probabilities will be calculated as below:

a.

[tex]P(X=x)=\frac{5^xe^{-5}}{x!}\\P(X\leq 8)=P(X=0)+P(X=1)+...+P(X=8)\\=\frac{5^0e^{-5}}{1!}+...+\frac{5^8e^{-5}}{8!}\\=0.006738+...+0.065278\\=0.961[/tex]

b.

[tex]P(X=8)=\frac{5^8e^{-5}}{8!}\\=0.065[/tex]

c.

[tex]P(9\leq X)=1-P(X=0)-P(X=1)-...-P(X=8)\\=0.068[/tex]

d.

[tex]P(5\leq X\leq 8)=P(X=5)+P(X=6)+P(X=7)+P(X=8)\\=0.491[/tex]

e.P(5<X<8)

[tex]P(X=6)+P(X=7)\\=0.251[/tex]

Answer:

16.35% probability their combined weight exceeds 46000 pounds.

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], the sample means with size n of at least 30 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 = 1135, \sigma = 97, n = 40, s = \frac{97}{\sqrt{40}} = 15.34[/tex]

Find the probability their combined weight exceeds 46000 pounds.

This is 1 subtracted by the pvalue of Z when [tex]X = \frac{46400}{40} = 1150[/tex]. So

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

By the Central Limit Theorem

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

[tex]Z = \frac{1150 - 1135}{15.34}[/tex]

[tex]Z = 0.98[/tex]

[tex]Z = 0.98[/tex] has a pvalue of 0.8365

1 - 0.8365 = 0.1635

16.35% probability their combined weight exceeds 46000 pounds.

Final answer:

The probability that the combined weight of 40 randomly selected one-year-old Hereford bulls exceeds 46,000 pounds is approximately 16.35%. This calculation uses the Z-score related to the sample mean weight exceeding 1,150 pounds.

Explanation:

To find the probability that the combined weight of 40 randomly selected one-year-old Hereford bulls exceeds 46,000 pounds, we first need to determine if the average weight per bull exceeds 1,150 pounds (since 46,000 ÷ 40 = 1,150). The mean weight of a Hereford bull is given as 1,135 pounds with a standard deviation of 97 pounds. Using the Central Limit Theorem, we can find the mean and standard deviation for the sample mean weight of 40 bulls.

The standard deviation of the sample mean (σ_x-bar) is σ ÷ √{n} = 97 pounds ÷ √{40}, which equals approximately 15.34 pounds. To find the Z-score, we use the formula Z = (X - μ) ÷ σ_x-bar, where X is 1,150 pounds, μ (the mean) is 1,135 pounds, and σ_x-bar is 15.34 pounds, yielding a Z-score of about 0.98.

Using the standard normal distribution table, a Z-score of 0.98 corresponds to a probability of approximately 0.8365. However, since we want the probability that the sample mean is greater than 1,150 pounds, we need to look at the upper tail, which is 1 - 0.8365 = 0.1635. Therefore, there's a 16.35% chance the combined weight of 40 randomly selected one-year-old Hereford bulls exceeds 46,000 pounds.

Answer:

Step-by-step explanation:

a) We would apply the periodic interest rate formula which is expressed as

P = a/[{(1+r)^n]-1}/{r(1+r)^n}]

Where

P represents the monthly payments.

a represents the amount of the loan

r represents the annual rate.

n represents number of monthly payments. Therefore

a = $22000

r = 0.065/12 = 0.0054

n = 12 × 3 = 36

Therefore,

P = 22000/[{(1+0.0054)^36]-1}/{0.0054(1+0.0054)^36}]

22000/[{(1.0054)^36]-1}/{0.0054(1.0054)^36}]

P = 22000/{1.214 -1}/[0.0054(1.214)]

P = 22000/(0.214/0.0065556)

P = 22000/32.64

P = $674

b) The total amount that he would pay in 3 years is

674 × 36 = 24265

total interest paid over the life of the loan is

24265 - 22000 = $2265

Answer:

[tex]z=\frac{0.45 -0.508}{\sqrt{\frac{0.508(1-0.508)}{50}}}=-0.82[/tex]  

[tex]p_v =P(z<-0.82)=0.206[/tex]  

So the p value obtained was a very high value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to  FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of newborn boys babies is not significantly lower than 0.508

Step-by-step explanation:

Data given and notation

n=50 represent the random sample taken

[tex]\hat p=0.45[/tex] estimated proportion of newborn boys babies

[tex]p_o=0.508[/tex] is the value that we want to test

[tex]\alpha=0.05[/tex] represent the significance level

Confidence=95% or 0.95  (asumed)

z would represent the statistic (variable of interest)

[tex]p_v[/tex] represent the p value (variable of interest)  

Concepts and formulas to use  

We assume that we want to check if the true proportion is less than 0.508.

Null hypothesis:[tex]p\geq 0.508[/tex]  

Alternative hypothesis:[tex]p < 0.508[/tex]  

When we conduct a proportion test we need to use the z statisitc, and the is given by:  

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

The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].

Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

[tex]z=\frac{0.45 -0.508}{\sqrt{\frac{0.508(1-0.508)}{50}}}=-0.82[/tex]  

Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level assumed is [tex]\alpha=0.05[/tex]. The next step would be calculate the p value for this test.  

Since is a left tailed test the p value would be:  

[tex]p_v =P(z<-0.82)=0.206[/tex]  

So the p value obtained was a very high value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to  FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of newborn boys babies is not significantly lower than 0.508

It is concluded that the null hypothesis H₀ is not rejected. There is not enough evidence to claim that the population proportion p is less than 0.508, at the α =0.05 significance level.

Suppose that we have:

Then, the z test statistic for one sample proportion is:

[tex]Z = \dfrac{\hat{p} - p_0}{\sqrt{\dfrac{p_0(1-p_0)}{n}}}[/tex]

Making the hypothesis and performing the test, assuming that we want to test if the population mean proportion is < 0.508 at the level of significance 0.05.

The following null and alternative hypotheses for the population proportion needs to be tested:

[tex]H_0: p \geq 0.508 \\\\ H_a: p & < & 0.508 \end{array}[/tex]

This corresponds to a left-tailed test, for which a z-test for one population proportion will be used.

Based on the information provided, the significance level is [tex]\alpha = 0.05[/tex], and the critical value for a left-tailed test is [tex]z_c = -1.64[/tex]

The rejection region for this left-tailed test is [tex]R = \{z: z < -1.645\}[/tex]

The z-statistic is computed as follows:

[tex]z & = & \displaystyle \frac{\hat p - p_0}{\sqrt{ \displaystyle\frac{p_0(1-p_0)}{n}}} \\\\& = & \displaystyle \frac{0.45 - 0.508}{\sqrt{ \displaystyle\frac{ 0.508(1 - 0.508)}{50}}} \\\\ & = & -0.82[/tex]

Since it is observed that [tex]z = -0.82 \ge z_c = -1.645[/tex],  it is then concluded that the null hypothesis is not rejected.

Thus, it is concluded that the null hypothesis H₀ is not rejected. Therefore, there is not enough evidence to claim that the population proportion p is less than 0.508, at the α =0.05 significance level.

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Answer:

[tex]P(\bar X >1650)=P(Z>\frac{1650-1637.52}{\frac{623.16}{\sqrt{400}}}=0.401)[/tex]

And we can use the complement rule and we got:

[tex]P(Z>0.401) =1-P(Z<0.401) = 1-0.656= 0.344[/tex]

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the bank account balances of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(1637.52,623.16)[/tex]  

Where [tex]\mu=1637.52[/tex] and [tex]\sigma=623.16[/tex]

Since the distribution of X is normal then the distribution for the sample mean [tex]\bar X[/tex] is given by:

[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]

And we can use the z score formula given by:

[tex] z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

And using this formula we got:

[tex]P(\bar X >1650)=P(Z>\frac{1650-1637.52}{\frac{623.16}{\sqrt{400}}}=0.401)[/tex]

And we can use the complement rule and we got:

[tex]P(Z>0.401) =1-P(Z<0.401) = 1-0.656= 0.344[/tex]

Answer: the probability is 0.49

Step-by-step explanation:

Since the account balances at the large bank are normally distributed.

we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = account balances.

µ = mean account balance.

σ = standard deviation

From the information given,

µ = $1,637.52

σ = $623.16

We want to find the probability that a simple random sample of 400 accounts has a mean that exceeds $1,650. It is expressed as

P(x > 1650) = 1 - P(x ≤ 1650)

For x = 1650,

z = (1650 - 1637.52)/623.16 = 0.02

Looking at the normal distribution table, the probability corresponding to the z score is 0.51

P(x > 1650) = 1 - 0.51 = 0.49

Answer:

0.6844 is the required probability.        

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = $1,250

Standard Deviation, σ = $125

We are given that the distribution of daily sales is a bell like shaped distribution that is a normal distribution.

Formula:

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

We have to find

P(sales less than $1,310)

[tex]P( x < 1310) = P( z < \displaystyle\frac{1310 - 1250}{125}) = P(z < 0.48)[/tex]

Calculation the value from standard normal z table, we have,  

[tex]P(x < 1310) =0.6844= 68.44\%[/tex]

0.6844 is the probability that sales on a given day at this store are less than $1,310.

The probability that sales on a given day at this store are less than $1,310 is [tex]68.44\%[/tex]

It is given that, mean [tex]\mu=1250[/tex] and deviation [tex]\sigma=125[/tex]

The z- score is given as,

                   [tex]z-score=\frac{x-\mu}{\sigma} \\\\z=\frac{1310-1250}{125}=0.48 \\\\[/tex]

We have to find probability that the sales on a given day at this store are less than $1,310

                [tex]P(x < 1310)=P(z < 0.48)[/tex]

From z- value table.

           [tex]P(x < 1310)=68.44\%[/tex]

The probability that sales on a given day at this store are less than $1,310 is [tex]68.44\%[/tex]

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Answer:

The Manufacturing cycle​ efficiency = 69 %

Step-by-step explanation:

The waiting time is  = 72 minute

The manufacturing lead time is ( [tex]T_{lead}[/tex] )= 160 minutes

Total cycle time ( [tex]T_{c}[/tex] ) = 160 + 72 = 232 minutes

Manufacturing cycle​ efficiency =  [tex]\frac{T_{lead} }{T_{c} }[/tex]

⇒ Manufacturing cycle​ efficiency = [tex]\frac{160}{232}[/tex]

⇒ Manufacturing cycle​ efficiency = 69 %

Therefore, the Manufacturing cycle​ efficiency = 69 %

The arc length of AB = 8.37 meters

Solution:

Degree of AB (θ) = 60°

Radius of the circle = 8 m

Let us find the arc length of AB.

Arc length formula:

[tex]$\text{Arc length}=2 \pi r\left(\frac{\theta}{360^\circ}\right)[/tex]

                 [tex]$=2 \times 3.14 \times 8 \left(\frac{60^\circ}{360^\circ}\right)[/tex]

                 [tex]$=2 \times 3.14 \times 8 \left(\frac{1}{6}\right)[/tex]

Arc length = 8.37 m

Hence the arc length of AB is 8.37 meters.

Final answer:

The standard deviation of the sample of numbers is approximately 10.68, after calculating the mean, finding the deviations, squaring them, averaging the squared deviations to get the variance, and then taking the square root of the variance.

Explanation:

The student is asking how to calculate the standard deviation of a sample of numbers. The sample given consists of the counts of bacteria found in ocean water, which are 41, 62, 45, 48, 69. To calculate the standard deviation, follow these steps:

Performing the calculations, you get:

Answer: there are 70 dimes and 40 quarters.

Step-by-step explanation:

A dime is 10 cents

A nickel is 5 cents

A quarter is 25 cents

Let x represent the number of dimes at the bottom.

Let y represent the number of quarters at the bottom.

The city park employee collected 1,850 cents in nickels, dimes, and quarters at the bottom of a wishing well. If there were 30 nickels, it means that

10x + 25y + 30 × 5 = 1850

10x + 25y = 1850 - 150

10x + 25y = 1700- - - - - - - - -1

There is a combined total of 110 dimes and quarters. It means that

x + y = 110

Substituting x = 110 - y into equation 1, it becomes

10(110 - y) + 25y = 1700

1100 - 10y + 25y = 1700

1100 + 15y = 1700

15y = 1700 - 1100

15y = 600

y = 600/15

y = 40

x = 110 - y = 110 - 40

x = 70

Answer:

Mean and standard deviation of the sample mean are 225.8 and 3.56 respectively.

Step-by-step explanation:

The mean (μₓ) and standard deviation of the sample mean (σₓ) are related to the mean (μ) and standard deviation of the population (σ) through the following relationship

μₓ = μ = 225.8

σₓ = σ/√n = 17.8/√25 = 17.8/5 = 3.56

Answer:

Step-by-step explanation:

Hello!

X: number of gypsy moths in a randomly selected trap.

This variable is strongly right-skewed. with a standard deviation of 1.4 moths/trap.

The mean number is 1.2 moths/trap, but several have more.

a.

The population is the number of moths found in traps places by the agriculture departments.

The population mean μ= 1.2 moths per trap

The population standard deviation δ= 1.4 moths per trap

b.

There was a random sample of 60 traps,

The sample mean obtained is X[bar]= 1

And the sample standard deviation is S= 2.4

c.

As the text says, this variable is strongly right-skewed, if it is so, then you would expect that the data obtained from the population will also be right-skewed.

d. and e.

Because you have a sample size of 60, you can apply the Central Limit Theorem and approximate the distribution of the sampling mean to normal:

X[bar]≈N(μ;σ²/n)

The mean of the distribution is μ= 1.2

And the standard deviation is σ/√n= 1.4/50= 0.028

f. and g.

Normally the distribution of the sample mean has the same shape of the distribution of the original study variable. If the sample size is large enough, as a rule, a sample of size greater than or equal to 30 is considered sufficient you can apply the theorem and approximate the distribution of the sample mean to normal.

You have a sample size of n=10 so it is most likely that the sample mean will have a right-skewed distribution as the study variable. The sample size is too small to use the Central Limit Theorem, that is why you cannot use the Z table to calculate the asked probability.

I hope it helps!

Answer:

[tex]z=\frac{0.18 -0.1}{\sqrt{\frac{0.1(1-0.1)}{100}}}=2.67[/tex]  

[tex]p_v =P(z>2.67)=0.0038[/tex]  

So the p value obtained was a very low value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of adults who rate Pepsi as being "concerned with my health" i significantly higher than 0.1

Step-by-step explanation:

Data given and notation

n=100 represent the random sample taken

X=18 represent the adults who rate Pepsi as being "concerned with my health"

[tex]\hat p=\frac{18}{100}=0.18[/tex] estimated proportion of adults who rate Pepsi as being "concerned with my health"

[tex]p_o=0.10[/tex] is the value that we want to test

[tex]\alpha=0.05[/tex] represent the significance level

z would represent the statistic (variable of interest)

[tex]p_v[/tex] represent the p value (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion is higher than 0.1.:  

Null hypothesis:[tex]p \leq 0.1[/tex]  

Alternative hypothesis:[tex]p > 0.1[/tex]  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

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

The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].

Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

[tex]z=\frac{0.18 -0.1}{\sqrt{\frac{0.1(1-0.1)}{100}}}=2.67[/tex]  

Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The next step would be calculate the p value for this test.  

Since is a right tailed test the p value would be:  

[tex]p_v =P(z>2.67)=0.0038[/tex]  

So the p value obtained was a very low value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of adults who rate Pepsi as being "concerned with my health" i significantly higher than 0.1

The student is asked to perform a hypothesis test at the  alpha = 0.05 level to determine if the percentage of participants rating Pepsi as 'concerned with my health' has statistically increased from 10% to 18% in a follow-up survey.

The question involves conducting a hypothesis test to determine if the percentage of participants rating Pepsi as being "concerned with my health" has increased after a new initiative. With an initial percentage of 10%, a follow-up survey showed that 18 out of 100 (or 18%) now rate Pepsi in this manner. To determine if this is a statistically significant increase, at the alpha = 0.05 significance level, a hypothesis test can be performed using the binomial or normal approximation to the binomial distribution.

Null hypothesis (H0): p = 0.10 (The percentage of those who believe Pepsi is concerned with their health has not changed)
Alternative hypothesis (Ha): p > 0.10 (The percentage has increased)

If the calculated p-value is less than the significance level (0.05), we can reject the null hypothesis in favor of the alternative hypothesis, thus concluding that the percentage has indeed increased.

The question given is incomplete, I googled and got the complete question as below:

You are a waterman daily plying the waters of Chesapeake Bay for blue crabs (Callinectes sapidus), the best-tasting crustacean in the world. Crab populations and commercial catch rates are highly variable, but the fishery is under constant pressure from over-fishing, habitat destruction, and pollution. These days, you tend to pull crab pots containing an average of 2.4 crabs per pot. Given that you are economically challenged as most commercial fishermen are, and have an expensive boat to pay off, you’re always interested in projecting your income for the day. At the end of one day, you calculate that you’ll need 7 legal-sized crabs in your last pot in order to break even for the day. Use these data to address the following questions. Show your work.

a. What is the probability that your last pot will have the necessary 7 crabs?

b. What is the probability that your last pot will be empty?

Answer:

a. Probability = 0.0083

b. Probability = 0.0907

Step-by-step explanation:

This is Poisson distribution with parameter λ=2.4

a)

The probability that your last pot will have the necessary 7 crabs is calculated below:

P(X=7)=  {e-2.4*2.47/7!} = 0.0083

b)

The probability that your last pot will be empty is calculated as:

P(X=0)=  {e-2.4*2.40/0!} = 0.0907

The probability of getting 7 crabs in the last pot is approximately 0.0082, while the probability of the last pot being empty is roughly 0.0907.

To solve this problem, we need to use the Poisson distribution. The Poisson distribution is a probability distribution that can be used to predict the number of events occurring within a fixed interval of time or space. Given that the average (λ) number of crabs per pot is 2.4, we can proceed to solve for the probabilities.

a. Probability of having 7 crabs in the last pot:

b. Probability of the last pot being empty:

The complete question is

You are a waterman daily plying the waters of Chesapeake Bay for blue crabs (Callinectes sapidus), the best-tasting crustacean in the world. Crab populations and commercial catch rates are highly variable, but the fishery is under constant pressure from overfishing, habitat destruction, and pollution. These days, you tend to pull crab pots containing an average of 2.4 crabs per pot. Given that you are economically challenged as most commercial fishermen are, and have an expensive boat to pay off, you’re always interested in projecting your income for the day. At the end of one day, you calculate that you’ll need 7 legal-sized crabs in your last pot in order to break even for the day

Use these data to address the following questions. Show your work.

a. What is the probability that your last pot will have the necessary 7 crabs?

b. What is the probability that your last pot will be empty?

The ODE is separable:

[tex]\dfrac{\mathrm dV}{\mathrm dt}=2\sqrt V\implies\dfrac{\mathrm dV}{2\sqrt V}=\mathrm dt[/tex]

Integrate both sides to get

[tex]\sqrt V=t+C[/tex]

Assuming the bubble starts with zero volume, so that [tex]V(0)=0[/tex], we find

[tex]\sqrt0=0+C\implies C=0[/tex]

Then the volume of the bubble at time [tex]t[/tex] is

[tex]V(t)=t^2[/tex]

and so the time it take for the bubble to reach a volume of 629 cubic cm is

[tex]t^2=529\implies t=23[/tex]

or 23 seconds after the teen first starts blowing the bubble.

Answer:

r(t) = (e^t +1, -cos(t) + 3, tan(t) + 2)

Step-by-step explanation:

A primitive of e^t is e^t+c, since r(0) has 2 in its first cooridnate, then

e^0+c = 2

1+c = 2

c = 1

Thus, the first coordinate of r(t) is e^t + 1.

A primitive of sin(t) is -cos(t) + c (remember that the derivate of cos(t) is -sin(t)). SInce r(0) in its second coordinate is 2, then

-cos(0)+c = 2

-1+c = 2

c = 3

Therefore, in the second coordinate r(t) is equal to -cos(t)+3.

Now, lets see the last coordinate.

A primitive of sec²(t) is tan(t)+c (you can check this by derivating tan(t) = sin(t)/cos(t) using the divition rule and the property that cos²(t)+sin²(t) = 1 for all t). Since in its third coordinate r(0) is also 2, then we have that

2 = tan(0)+c = sin(0)/cos(0) + c = 0/1 + c = 0

Thus, c = 2

As a consecuence, the third coordinate of r(t) is tan(t) + 2.

As a result, r(t) = (e^t +1, -cos(t) + 3, tan(t) + 2).

To find the function r(t) that satisfies the given condition, integrate each component of r'(t) and solve for the constants using the initial condition.

To find the function r(t) that satisfies the condition r'(t) = (e^t, sin t, sec^2 t) and r(0) = (2, 2, 2), we need to integrate each component of r'(t) with respect to t.

Combining these integrals, we get r(t) = (e^t + C_1, -cos t + C_2, tan t + C_3). Substituting the initial condition r(0) = (2, 2, 2), we can solve for the constants and obtain the function r(t) = (e^t + 2, -cos t + 2, tan t + 2).

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The probability that all the answers are the same in the section of the exam is 1/8 or 12.5%.

To find the probability that all the answers are the same, we need to consider the number of ways in which all the answers can be the same, divided by the total number of possible outcomes. Since there are only two possible answers (True or False) for each question, there are a total of 2^4 = 16 possible outcomes.

Out of these 16 outcomes, there are only two ways in which all the answers can be the same: either all True or all False.

Therefore, the probability that all the answers are the same is 2/16 = 1/8 = 0.125, or 12.5%.

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Answer:

-b/a and line is falling

Step-by-step explanation:

Let (-a,0) and (0,-b) be point 1 and 2, respectively. The slope passing through point (-a,0) and (0,-b) can be calculated using the following equation

[tex]m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-b - 0}{0 - (-a)} = -b/a[/tex]

As b and a are positive real number, then the slope m = -b/a is a negative real number. The line passing through these 2 points is falling.

The slope of the line passing through points (-a, 0) and (0, -b) is calculated using the formula for slope, resulting in a slope of -b/a, which indicates that the line falls as it moves from left to right.

To calculate the slope (m), we use the formula m = (y2 - y1) / (x2 - x1), with (x1, y1) = (-a, 0) and (x2, y2) = (0, -b). Substituting the values, we get m = (-b - 0) / (0 - (-a)) = -b/a. Since both a and b are positive real numbers, the sign of the slope depends on the negative numerator, indicating the line falls as it goes from left to right. If a and/or b were zero, revealing a vertical or horizontal line, a different methodology would apply, such as indicating an undefined slope for a vertical line or a zero slope for a horizontal line.

Answer:

405150

Step-by-step explanation:

3 people are to be selected from 75 people. This could be done using the combination function, [tex]\binom{75}{3}[/tex]. But thus function only gives the number of ways of choosing any 3 out of the 75 customers. For any single selection of any 3 customers, there are 3! ways of sharing the three prizes among them. Hence the total number of ways of sharing the prizes is

[tex]\binom{75}{3}\times3! = 405150[/tex]

In fact, this is the permutation function given by

[tex]{}^{75}P_3 = \dfrac{75!}{(75-3)! = 405150}[/tex]

The differential equations for x and y are:

x = (cost - D^2 (t2 + 3t)) / ((D+5) + D^2 (D+9))

y = (t2 + 3t - D2 cost) / (D2 / (D+5) + (D+9))

Here's how to solve the problem:

(a) Write the system as a differential equation for x:

Eliminate y: Substitute the second equation into the first equation to eliminate y.

(D+5)x + D2(t2 + 3t) = cost

Rearrange and define operator: Define P(D) = (D+5) + D^2 (D+9).

P(D) x = cost - D^2 (t2 + 3t)

Write as differential equation:

x = (cost - D^2 (t2 + 3t)) / P(D)

(b) Write the system as a differential equation for y:

Eliminate x: Substitute the first equation into the second equation to eliminate x.

D2((cost - D2y) / (D+5)) + (D+9)y = t2 + 3t

Rearrange and define operator: Define Q(D) = D2 / (D+5) + (D+9).

Q(D) y = t2 + 3t - D2 cost

Write as differential equation:

y = (t2 + 3t - D2 cost) / Q(D)

Therefore, the differential equations for x and y are:

x = (cost - D^2 (t2 + 3t)) / ((D+5) + D^2 (D+9))

y = (t2 + 3t - D2 cost) / (D2 / (D+5) + (D+9))

(a)[tex]\(x = \frac{1}{D^3 + 14D + 5}\cdot \left( \cos(t) \right)\)[/tex]

(b)[tex]\(y = \frac{1}{D^3 + 14D + 5}\cdot \left( t^2 + 3t \right)\)[/tex]   by expressing (x) and (y) in terms of differential operators and given functions of (t), we derive their differential equations .

The given system of equations can be transformed into differential equations using operator (D), representing differentiation with respect to a variable, often time.

(a) To derive the differential equation for (x), we isolate (x) in terms of the differential operator \(D\). Rearranging the first equation gives us [tex]\(x = \frac{\cos(t)}{D+5} - \frac{D^2y}{D+5}\).[/tex]Substituting the expression for (y) from the second equation into this yields[tex]\(x = \frac{\cos(t)}{D+5} - \frac{D^2}{D+5}\left(\frac{t^2 + 3t}{D+9}\right)\).[/tex] Simplifying further leads to [tex]\(x = \frac{1}{D^3 + 14D + 5}\cdot \left( \cos(t) \right)[/tex]\).

(b) Similarly, isolating \(y\) from the given equations leads to [tex]\(y = \frac{t^2 + 3t}{D+9} - \frac{D+5}{D+9}x\).[/tex]Substituting the expression for (x) from the first equation into this yields [tex]\(y = \frac{t^2 + 3t}{D+9} - \frac{D+5}{D+9}\left(\frac{\cos(t)}{D+5} - \frac{D^2y}{D+5}\right)\).[/tex]Further simplification results in [tex]\(y = \frac{1}{D^3 + 14D + 5}\cdot \left( t^2 + 3t \right)\).[/tex]

Therefore, by expressing (x) and (y) in terms of differential operators and given functions of (t), we derive their differential equations as specified in the final answer.

Answer:

A hypothesis is a proposed statement explaining the relationship between two or more variable which can be verifiable through experiment. In this case a testable hypothesis would be.

When yeast and sugar are mixed together during formation, it will cause the sugar to undergo a high rate of glycolysis and CO[tex]^{2}[/tex] will be present in the fermentation tube.

Answer:

[tex] 28x^2 [/tex]

Step-by-step explanation:

[tex]area \:o f \: \triangle = \frac{1}{2} \times 8x \times 7x \\ = 4x \times 7x \\ = 28 {x}^{2} [/tex]

Answer:

c. Inductive and Strong

Step-by-step explanation:

In inductive reasoning, provided data is analyzed in order to reach a conclusion. In this case, the argument provides data regarding Jane and Nancy's awards and their love for mathematics and then draws a conclusion regarding Nancy's performance in a particular class, this is an example of inductive reasoning.

As for the strength of the argument, it is plausible to infer that Jane and Nancy have similar mathematics skills since they both love calculus and excel academically. Therefore, if Jane does well in the calculus class, it is a strong argument to say that Nancy does as well.

The answer is :

c. Inductive and Strong

Final answer:

The argument is Deductive and Invalid as it does not guarantee Nancy's success in the calculus class based on Jane's performance.

Explanation:

The argument presented is Deductive and Invalid. This is because the conclusion that Nancy will do well in the calculus class based on Jane's performance does not necessarily follow logically. Just because Jane excelled does not guarantee the same result for Nancy.

In deductive reasoning, the conclusion must follow necessarily from the premises, which is not the case here. An example of a valid deductive argument would be 'All humans are mortal, Socrates is a human, therefore Socrates is mortal.'

Because the argument in question does not provide a guarantee of Nancy's success based on Jane's performance, it is categorized as Invalid in the realm of deductive reasoning.

Let x* denote Peter's score. Then

P(X > x*) = 0.025

P((X - 896)/174 > (x* - 896)/174) = 0.025

P(Z > z*) = 1 - P(Z < z*) = 0.025

P(Z < z*) = 0.975

where Z follows the standard normal distribution (mean 0 and std dev 1).

Using the inverse CDF, we find

P(Z < z*) = 0.975  ==>  z* = 1.96

Then solve for x*:

(x* - 896)/174 = 1.96  ==>  x* = 1237.04

so Peter's score is roughly 1237.

Answer:

We conclude that there is not enough evidence to support the claim that company is spending too much time on email.

Step-by-step explanation:

We are given the following in the question:  

Population mean, μ = 96 minutes

Sample mean, [tex]\bar{x}[/tex] = 91 minutes

Sample size, n = 100

Alpha, α = 0.05

Sample standard deviation, s = 25 minutes

First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu \leq 96\text{ minutes}\\H_A: \mu > 96\text{ minutes}[/tex]

We use one-tailed t test to perform this hypothesis.

Formula:

[tex]t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }[/tex]

Putting all the values, we have

[tex]t_{stat} = \displaystyle\frac{91 - 96}{\frac{25}{\sqrt{100}} } = -2[/tex]

Now, [tex]t_{critical} \text{ at 0.05 level of significance, 99 degree of freedom } = 1.6603[/tex]

Since,                        

The calculated test statistic is less than the critical value, we fail to reject the null hypothesis and accept it.

Thus, we conclude that there is not enough evidence to support the claim that company is spending too much time on email.

Answer:

11.51% probability that the sample average will be less than 182

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 = 187, \sigma = 32, n = 64, s = \frac{32}{\sqrt{64}} = 4[/tex]

What is the probability that the sample average will be less than 182

This is the pvalue of Z when X = 182. So

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

By the Central Limit Theorem

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

[tex]Z = \frac{182 - 187}{4}[/tex]

[tex]Z = -1.2[/tex]

[tex]Z = -1.2[/tex] has a pvalue of 0.1151

11.51% probability that the sample average will be less than 182