A builder of houses needs to order some supplies that have a waiting time Y for delivery, with a continuous uniform distribution over the interval from 1 to 4 days. Because she can get by without them for 2 days, the cost of the delay is fixed at $400 for any waiting time up to 2 days. After 2 days, however, the cost of the delay is $400 plus $50 per day (prorated) for each additional day. That is, if the waiting time is 3.5 days, the cost of the delay is $400 $50(1.5)
Find the expected value of the builder’s cost due to waiting for supplies.

Answer:

Step-by-step explanation:

Hello!

X: nitrate concentration in the city's drinking water supply.

X~N(μ;δ²)

A concentration above 10 ppm is a health risk for infants less than 6 months old.

The reported nitrate concentration is between 1.59 and 2.52 ppm, values outside this range are considered unusual.

If any value of a normal distribution that is μ ± 2δ is considered unusual, it is determined that:

μ - 2δ= 1.59

μ + 2δ= 2.52

20) and 21)

I'll clear the values of the mean and the standard deviation using the given information:

a) μ - 2δ= 1.59 ⇒ μ = 1.59 + 2δ

b) μ + 2δ= 2.52 ⇒ Replace the value of Mu from "a" in "b" and clear the standard deviation:

(1.59 + 2δ) + 2δ= 2.52

1.59 + 4δ= 2.52

4δ= 2.52 - 1.59

δ= 0.93/4

δ= 0.2325

Then the value of the mean is μ = 1.59 + 2(0.2325)

μ = 1.59 + 0.465

μ = 2.055

22)

Acording to the empirical rule, 68% of a normal distribution is between μ±δ

Then you can expect 68% of the distribution between:

μ-δ= 2.055-0.2325= 1.8225

μ+δ= 2.055+0.2325= 2.2875

Correct option: C. 1.823 and 2.288 ppm

I hope it helps!

Answer:

P(X = 17) = 0.3002

Step-by-step explanation:

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

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

In which [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]

And p is the probability of X happening.

In this problem we have that:

[tex]n = 18, p = 0.9[/tex]

We want P(X = 17). So

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

[tex]P(X = 17) = C_{18,17}.(0.9)^{17}.(0.1)^{1} = 0.3002[/tex]

P(X = 17) = 0.3002

Answer:

P(X=17) = 0.3002 .

Step-by-step explanation:

We are given that the random variable X has a binomial distribution with the given probability of obtaining a success.

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 = 18

            r = number of success = 17

           p = probability of success which in our question is given as 0.9 .

So, X ~ [tex]Binom(n=18,p=0.9)[/tex]

We have to find the probability of P(X = 17);

P(X = 17) = [tex]\binom{18}{17}0.9^{17} (1-0.9)^{18-17}[/tex]

               = [tex]18 \times 0.9^{17} \times 0.1^{1}[/tex]    { [tex]\because \binom{n}{r} = \frac{n!}{r! \times (n-r)!}[/tex] }

               = 0.3002

Therefore, P(X=17) = 0.3002 .

Answer:

There is enough evidence to support the claim that students study more than two hours per weeknight on the average.      

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 2 hours = 120 minutes

Sample mean, [tex]\bar{x}[/tex] = 148 minutes = 2.467 hours

Sample size, n = 255

Alpha, α = 0.05

Population standard deviation, σ = 65 minutes

First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu = 120\text{ minutes}\\H_A: \mu > 120\text{ minutes}[/tex]

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

Formula:

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

Putting all the values, we have

[tex]z_{stat} = \displaystyle\frac{148 - 120}{\frac{65}{\sqrt{255}} } = 6.8788[/tex]

Now, we calculate the p-value from the normal standard table.

P-value = 0.00001

Since the p-value is smaller than the significance level, we fail accept the null hypothesis and reject it, We accept the alternate hypothesis.

Conclusion:

Thus, there is enough evidence to support the claim that students study more than two hours per weeknight on the average.

Answer:

There is enough evidence to support the claim that students study more than two hours per weeknight on the average.    

Step-by-step explanation:

Answer:

B. approximately normal with a mean of 2.02 dollars and a standard error of 0.45 dollars

Step-by-step explanation:

We use the Central Limit Theorem to solve this question.

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

In this problem, we have that:

[tex]\mu = 2.02, \sigma = 3, n = 45, s = \frac{3}{\sqrt{45}} = 0.45[/tex]

So the correct answer is:

B. approximately normal with a mean of 2.02 dollars and a standard error of 0.45 dollars

Answer:

a. Should NOT be rejected because all the parameters are generally acceptable.

Step-by-step explanation:

Answer:

1) Number of scoops = 42

2)Number of scoops = 10

Step-by-step explanation:

Volume of the sink = (4000/3)π in³

1)

Diameter = 4 in      ∵   Radius = 4/2 = 2 in

Height = 8 in

Volume of the cylindrical cup =  [tex]\pi r^{2}h = \pi 2^{2} * 8 = 32\pi[/tex] in³

Number of cups of water that must be scooped out = Volume of the sink/ Volume of the cylindrical cup

= (4000/3)π / 32π    ==>  42

Number of scoops = 42

b)

Diameter = 8 in    ∵   Radius = 8/2 = 4 in

Height = 8 in

Volume of the cylindrical cup =  [tex]\pi r^{2}h = \pi 4^{2} * 8 = 128\pi[/tex] in³

Number of cups of water that must be scooped out = Volume of the sink/ Volume of the cylindrical cup

= (4000/3)π / 128π ==> 10

Number of scoops = 10

Answer:

(b) 15.50 to 16.50 ounces

Step-by-step explanation:

The Empirical Rule states that, for a normally distributed(bell shaped) random variable:

68% of the measures are within 1 standard deviation of the mean.

95% of the measures are within 2 standard deviation of the mean.

99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we have that:

Mean = 16

Standard deviation = 0.25

Based on this information, between what two values could we expect 95% of all cans to weigh?

By the Empirical Rule, within 2 standard deviations of the mean. So

16 - 2*0.25 = 15.50 ounces

16 + 2*0.25 = 16.50 ounces

So the corret answer is:

(b) 15.50 to 16.50 ounces

Final answer:

By calculating two standard deviations from the mean in both directions, we determine that 95% of cans are expected to weigh between 15.5 and 16.5 ounces. Thus, option b is correct.

Explanation:

The question involves finding between which two values 95% of all cans of tomato soup would weigh, given that the mean weight is 16 ounces with a standard deviation of 0.25 ounces. This question can be answered by applying the empirical rule (also known as the 68-95-99.7 rule) for a normal distribution, which states that approximately 95% of the data falls within two standard deviations of the mean.

To calculate the range:

1. Subtract two standard deviations from the mean to find the lower limit.

16 ounces - (2 * 0.25 ounces) = 15.5 ounces

2. Add two standard deviations to the mean to find the upper limit.

16 ounces + (2 * 0.25 ounces) = 16.5 ounces

Therefore, we can expect 95% of all cans to weigh between 15.5 and 16.5 ounces, which corresponds to option (b) 15.50 to 16.50 ounces.

The second and third function are missing and they are;

The second function; ((2^n) + n^(2))(n^(3) + 3^(n))

The third function is;

(n^(n) + n^(2n) + 5^(n))(n! + 5^(n))

Answer:

A) O((n^(3)) logn)

B) O(6^(n))

C) O(n^(n)(n!))

Step-by-step explanation:

I've attached explanation of the 3 answers.

The big-O estimate for the function f(n) = [tex]f(n) = (n log n + n^2)(n^3 + 2)[/tex] is [tex]O(n^5)[/tex] , as the product of the highest order terms [tex]n^2[/tex] and [tex]n^3[/tex] is [tex]n^5[/tex], which dominates the function's growth.

To find a big-O estimate for the function [tex]f(n) = (n log n + n^2)(n^3 + 2)[/tex], we must look for the term with the highest growth rate as n goes to infinity. This function is the product of two terms, n log [tex]n log n + n^2[/tex] and [tex]n^3 + 2[/tex]. The term with the highest growth rate in the first part is n2, and in the second part, it is n3. Therefore, the product of the two highest order terms will give us the term that grows the fastest: [tex]n^2 \times n^3 = n^5[/tex].

Thus, the given function is in big-O of [tex]n^5[/tex], which means [tex]f(n) \in O(n^5)[/tex]. The higher order term [tex]n^5[/tex]completely dominates the growth of the function, making the other terms and constants irrelevant for big-O notation. Remember that big-O notation provides an upper bound and is used to describe the worst-case scenario for the growth rate of the function.

Learn more about big-O here:

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Answer:

See the attached picture.

Step-by-step explanation:

See the attached picture.

To ensure compositions f ∘ g, g ∘ f, and f ∘ f make sense, certain conditions must be met regarding the number of inputs and outputs of the functions.

(a) For f ∘ g to make sense, the number of inputs of g must match the number of outputs of f. Therefore, n must equal p, and m must equal q.

(b) For g ∘ f to make sense, the number of inputs of f must match the number of outputs of g. Therefore, n must equal q, and m must equal p.

(c) For f ∘ f to make sense, the number of outputs of f must match the number of inputs of f. Therefore, n must equal m.

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Answer:

1) 0.75

2) No, the 2 events are not mutually exclusive because P(A&B) ≠ 0

3)The 2 events are not independent because P(A | B) ≠ P(A).

Step-by-step explanation:

1) P(A) = 0.59 and P(B) = 0.41

Thus; to find the probability that at least one of the stocks will rise in price;

P(A or B) = P(A) + P(B) - P(A&B)

Now, we know that if B rises in price, the probability that A will also rise is 0.61 i.e P(A|B) = 0.61

Hence;P(A & B) = P(A|B) x P(B) =0.61 x 0. 41 = 0.25

So P(A or B) = 0.59 + 0.41 - 0.25 = 0.75

B) we want to find out if events A and B are mutually exclusive ;

P(A|B) = P(A&B)/P(B), For the events to be mutually exclusive, P(A&B) has to be zero. Since P(A&B) ≠ 0,the 2 events are not mutually exclusive.

C) Since P(A|B) = 0.61 and P(A) is 0.59. Thus, P(A | B) ≠ P(A).

So, therefore the 2 events are not independent.

Answer:

P(0.29a < X < 0.34a) = -(4a/π) [(cos 0.34πa²) - (cos 0.29πa²)]

Assuming the box is of unit length, a = 1,

0.29a = 0.29 and 0.34a = 0.34

P(0.29 < X < 0.34) = (-4/π) [(cos 0.34π) - (cos 0.29π)] = 0.167 = 0.17 to 2 s.f

Step-by-step explanation:

ψ(x) = 2a(√sin(πxa)

The probability of finding a particle in a specific position for a given energy level in a one-dimensional box is related to the square of the wavefunction

The probability of finding a particle between two points 0.29a and 0.34a is given mathematically as

P(0.29a < X < 0.34a) = ∫⁰•³⁴ᵃ₀.₂₉ₐ ψ²(x) dx

That is, integrating from 0.29a to 0.34a

ψ²(x) = [2a(√sin(πxa)]² = 4a² sin(πxa)

∫⁰•³⁴ᵃ₀.₂₉ₐ ψ²(x) dx = ∫⁰•³⁴ᵃ₀.₂₉ₐ (4a² sin(πxa)) dx

∫⁰•³⁴ᵃ₀.₂₉ₐ (4a² sin(πxa))

= - [(4a²/πa)cos(πxa)]⁰•³⁴ᵃ₀.₂₉ₐ

- [(4a/π)cos(πxa)]⁰•³⁴ᵃ₀.₂₉ₐ = -(4a/π) [(cos 0.34πa²) - (cos 0.29πa²)]

P(0.29a < X < 0.34a) = -(4a/π) [(cos 0.34πa²) - (cos 0.29πa²)]

Assuming the box is of unit length, a = 1

P(0.29 < X < 0.34) = (-4/π) [(cos 0.34π) - (cos 0.29π)] = (-1.273) [0.482 - 0.613] = 0.167.

Answer:

100% probability that between 1200 and 1450 of those surveyed incurred debt.

Step-by-step explanation:

We use the binomial approximation to the normal to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

[tex]E(X) = np[/tex]

The standard deviation of the binomial distribution is:

[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]

Normal probability distribution

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.

When we are approximating a binomial distribution to a normal one, we have that [tex]\mu = E(X)[/tex], [tex]\sigma = \sqrt{V(X)}[/tex].

In this problem, we have that:

[tex]n = 2500, p = 0.52[/tex]

So

[tex]\mu = E(X) = 2500*0.52 = 1300[/tex]

[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{2500*0.52*0.48} = 24.98[/tex]

Determine the probability that between 1200 and 1450 of those surveyed incurred debt.

This is the pvalue of Z when X = 1450 subtracted by the pvalue of Z when X = 1200. So

X = 1450

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

[tex]Z = \frac{1450 - 1300}{24.98}[/tex]

[tex]Z = 6[/tex]

[tex]Z = 6[/tex] has a pvalue of 1

X = 1200

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

[tex]Z = \frac{1200 - 1300}{24.98}[/tex]

[tex]Z = -4[/tex]

[tex]Z = -4[/tex] has a pvalue of 0

1 - 0 = 1

100% probability that between 1200 and 1450 of those surveyed incurred debt.

Answer:

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Step-by-step explanation:

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Answer:

5.55% of broilers weigh between 1143 and 1242 grams

Step-by-step explanation:

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.

In this problem, we have that:

[tex]\mu = 1511, \sigma = 198[/tex]

What proportion of broilers weigh between 1143 and 1242 grams?

This is the pvalue of Z when X = 1242 subtracted by the pvalue of Z when X = 1143.

X = 1242

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

[tex]Z = \frac{1242 - 1511}{198}[/tex]

[tex]Z = -1.36[/tex]

[tex]Z = -1.36[/tex] has a pvalue of 0.0869

X = 1143

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

[tex]Z = \frac{1143 - 1511}{198}[/tex]

[tex]Z = -1.36[/tex]

[tex]Z = -1.86[/tex] has a pvalue of 0.0314

0.0869 - 0.0314 = 0.0555

5.55% of broilers weigh between 1143 and 1242 grams

Answer:

(5.216) mean = 61.71

(5.217) standard deviation = 8.88

(5.218) P(X>=60) = 0.9238

(5.129) L = 79, U = 69

(5.130) P(X> or = U) = 0.7058

Step-by-step explanation:

The table of the statistic is set up as shown in attachment.

(5.216) mean = summation of all X ÷ no of data.

mean = 432/7 = 61.71 birds

(5.217)Standard deviation = √ sum of the absolute value of difference of X from mean ÷ number of data

S = √ /X - mean/ ÷ 7

= √551.428/7

S = 8.88

(5.218) P (X> or = 60)

= P(Z> or =60 - 61.71/8.8 )

= P(Z>or= - 0.192)

= 1 - P(Z< or = 0.192)

= 1- 0.0762

= 0.9238

(5.219)the 15th percentile=15/100 × 7

15th percentile = 1.05

The value is the number in the first position and that is 79,

L= 79

85th percentile = 85/100 × 7 = 5.95

The value is the number in the 6th position, and that is 69

U = 69

5.130) P(X>or = 60)

= P(Z>or= 69 - 61.71/8.8)

= P(Z> or = 0.8208)

= 1 - P(Z< or = 0.8209)

= 1 - 0.2942

= 0.7058

Answer:

The answer is c. The expected value.

Step-by-step explanation:

Final answer:

The weighted average of the value of a random variable calculated with probability weights is known as the expected value. The expected value indicates the long-term average outcome for a random variable that has been experimented on multiple times. It is calculated using the sum of the products of the variable values and their corresponding probabilities.

Explanation:

The weighted average of the value of a random variable, where the probability function provides weights, is known as the expected value. The expected value is a key concept in probability and statistics, often referred to in equations as E(X) and symbolized as μ (mu), where X is a random variable. It represents the long-term average outcome of a random variable after many repetitions of an experiment. For a discrete random variable with a probability distribution function P(x), the expected value can be calculated using the formula: μ = Σ XP(x).

Random variables are typically defined in the context of a probability distribution, which can take on various forms like normal, uniform, or exponential distributions, depending on the nature of the experiment. When constructing a confidence interval for a population mean, the confidence level indicates the degree of certainty in the interval estimate. The error bound, which is part of the confidence interval calculation, would typically decrease if the confidence level is lowered because a lower confidence level means that we are willing to accept a greater chance that the interval does not contain the population mean, leading to a narrower interval.

The question discusses two principal sources for conducting learning research: 1. nature of knowledge and diverse learning methods, 2. cognitive processes. The outlined chapters provide a systematic framework to understand personal learning, covering areas from motivation, mindset, individual learning styles to personality types, and practical application of learning knowledge.

The student's question pertains to the two fundamental sources for conducting research on learning. The first source refers to the nature of knowledge and the diversity of learning methods, while the second digs into cognitive processes or what goes on in our minds.

The chapters provided, under the title of 'Knowing Yourself as a Learner', present a systematic examination of personal learning. It starts with 'The Power to Learn', which considers the capacity and eagerness for learning. The 'Motivated Learner' and 'It's All in the Mindset' chapters elaborate on how motivation and mindset affect learning.

The 'Learning Styles' and 'Personality Types and Learning' sections delve into the individual differences that shape how we absorb and process information. Finally, 'Applying What You Know about Learning' encourages students to utilize their understanding of their learning attributes to optimize knowledge acquisition.

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Answer:

a) We have the following distribution [tex] X \sim N(\mu =650, \sigma =10)[/tex]

And we want to calculate:

[tex] P(X>654) [/tex]

And in order to calclate this in the ti 84 we can use the following code

2nd> Vars>DISTR

And then we need to use the following code:

1-normalcdf(-1000,654,650,10)

And we got:

[tex] P(X>654)=0.3445783 [/tex]

b) For this case we assume a normal standard distribution and we want to calculate:

[tex] P(z<0.72)[/tex]

And using the following code in the Ti84 or using the normal standard table we got:

normalcdf(-1000,0.72,0,1)

[tex] P(z<0.72)=0.76424[/tex]

So this part was the wrong solution from the solution posted,

Step-by-step explanation:

Part a

We have the following distribution [tex] X \sim N(\mu =650, \sigma =10)[/tex]

And we want to calculate:

[tex] P(X>654) [/tex]

And in order to calclate this in the ti 84 we can use the following code

2nd> Vars>DISTR

And then we need to use the following code:

1-normalcdf(-1000,654,650,10)

And we got:

[tex] P(X>654)=0.3445783 [/tex]

Part b

For this case we assume a normal standard distribution and we want to calculate:

[tex] P(z<0.72)[/tex]

And using the following code in the Ti84 or using the normal standard table we got:

normalcdf(-1000,0.72,0,1)

[tex] P(z<0.72)=0.76424[/tex]

So this part was the wrong solution from the solution posted,

The corrected parts of the question has been attached to this answer.

Answer:

A) Probability that the error is less than 0.2 mm; P(X < 0.2) = 0.0272

B) Mean Error (E(X)) = 0.6

C) Variance Error (V(X)) = 0.04

D) Answer properly written in attachment (Page 2)

E) P(0<X<0.8) = 0.8192

Step-by-step explanation:

The probability density function of X is;

f(x) = { 12(x^(2) −x^(3) ; 0<x<1

So, due to the integral symbol and for clarity sake, i have attached all the explanations for answers A - D.

E) The probability that the specification for the error to be between 0 to 0.8 mm is met will be;

P(0<X<0.8) = F(0.8) − F(0) =12([(0.8)^(3)] /3] −[(0.8)^(4)]/4]

= 0.8192

So, the probability is 0.8192.

Answer:

"half of the 12-inch quesadilla" is larger

Step-by-step explanation:

The "inch size" is the diameter of the circular quesadilla.

The area of circle is given by  [tex]\pi r^2[/tex]

We also know radius is half of diameter.

HALF OF 12-INCH QUESADILLA:

12in diameter = 6 inch radius, the area is:

[tex]\pi(6)^2=36\pi[/tex]

Half of this is:

[tex]\frac{36\pi}{2}=18\pi[/tex]

ENTIRE 6-INCH QUESADILLA:

6 inch diameter means 6/2 = 3 inch radius

The area of this is:

[tex]\pi(3)^2 = 9\pi[/tex]

So, we can say that "half of the 12-inch quesadilla" is larger.

The probability that the sum of the dots on two dice will be 3 is 0.056, determined by finding the number of favorable outcomes (2) and dividing by the total number of outcomes (36).

The subject of this question is probability, specifically focusing on an experiment involving the tossing of two six-sided dice. To find the probability that the sum of the dots on the two top faces will be 3, we first need to determine the total number of possible outcomes when two dice are thrown. For one six-sided die, there are 6 possible outcomes. Therefore, when two dice are thrown, the total number of outcomes is 6 * 6 = 36.

Next, we need to find the number of outcomes where the sum of the dots would be 3. Looking at our dice, we can see that this can only happen in two ways: getting a 1 on the first die and a 2 on the second, or getting a 2 on the first die and a 1 on the second.

So, the number of favorable outcomes is 2.

Probability is defined as the number of favorable outcomes divided by the total number of outcomes. Hence, the probability of the sum of the dots being 3 when two dice are thrown is 2/36. To get this in three significant figures, we divide 2 by 36 which gives the decimal 0.0556. Therefore, the probability to three significant figures is 0.056.

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Answer:

0.0497 is the required probability.

Step-by-step explanation:

We are given the following information in the question:

Mean number of calls = 3 calls per minute

The number of calls coming per minute is following a Poisson distribution.

Formula:

[tex]P(X =k) = \displaystyle\frac{\lambda^k e^{-\lambda}}{k!}\\\\ \lambda \text{ is the mean of the distribution}[/tex]

We have to evaluate

[tex]P( x =0) = \displaystyle\frac{\lambda^k e^{-\lambda}}{k!}\\\\P(x = 0) = \frac{3^0e^{-3}}{0!} = 0.0497[/tex]

0.0497 is the probability that no calls are received in a given one-minute period.

Answer:

32.33% probability of having at least 3 erros in an hour.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]

In which

x is the number of sucesses

e = 2.71828 is the Euler number

[tex]\mu[/tex] is the mean in the given time interval.

The mean number of errors is 2 per hour.

This means that [tex]\mu = 2[/tex]

(a) What is the probability of having at least 3 errors in an hour?

Either you have 2 or less errors in an hour, or we have at least 3 errors. The sum of the probabilities of these events is decimal 1. So

[tex]P(X \leq 2) + P(X \geq 3) = 1[/tex]

We want [tex]P(X \geq 3)[/tex]

So

[tex]P(X \geq 3) = 1 - P(X \leq 2)[/tex]

In which

[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)[/tex]

[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]

[tex]P(X = 0) = \frac{e^{-2}*(2)^{0}}{(0)!} = 0.1353[/tex]

[tex]P(X = 1) = \frac{e^{-2}*(2)^{1}}{(1)!} = 0.2707[/tex]

[tex]P(X = 2) = \frac{e^{-2}*(2)^{2}}{(2)!} = 0.2707[/tex]

[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.1353 + 0.2707 + 0.2707 = 0.6767[/tex]

[tex]P(X \geq 3) = 1 - P(X \leq 2) = 1 - 0.6767 = 0.3233[/tex]

32.33% probability of having at least 3 erros in an hour.

Final answer:

The question involves calculating the probability of having at least 3 transmission errors in an hour for a long-range radio using the Poisson distribution. Given the mean number of errors is 2 per hour, the resulting probability is found to be 32.33%.

Explanation:

The question involves understanding the Poisson distribution, which is a probability distribution that is used to describe the number of events occurring within a fixed interval of time or space given a known average rate. In this question, we are dealing with a long-range radio transmits across the Atlantic Ocean, where the average number of transmission errors per hour follows a Poisson distribution with a mean (λ) of 2 errors per hour. To find the probability of having at least 3 errors in any given hour, we can use the formula P(X ≥ k) = 1 - [P(X=0) + P(X=1) + P(X=2)], where X is the number of transmission errors.

To calculate P(X=k) for k=0,1,2, we use the Poisson probability formula P(X=k) = (e^{-λ} * λ^k) / k!, where e is the base of the natural logarithm (approximately 2.71828). Thus:

P(X=0) = (e^{-2} x 2⁰) / 0! = 0.1353

P(X=1) = (e^{-2} x 2¹) / 1! = 0.2707

P(X=2) = (e^{-2} x 2²) / 2! = 0.2707

Adding up these probabilities gives us 0.6767. Therefore, the probability of having at least 3 errors in any given hour is 1 - 0.6767 = 0.3233 or 32.33%.

Answer:

Yes, this a biased sampling method.

Step-by-step explanation:

The sampling done is a biased sampling method.

This is because usually people who are upset or dissatisfied with the service are the ones who write online reviews, as there are no other way to release their frustration with the service provided.

The bias is likely to be directed towards the proportion of people who are not satisfied with service.

Answer:

6 extra hot dogs.

Step-by-step explanation:

A dependent variable is the variable being tested and measured in a scientific experiment. The dependent variable is 'dependent' on the independent variable

The 6 hot dogs are dependent on whether the teammates will bring friends. In other words, if there are no friends, there are no hot dogs. Note that the teammates are the independent variable that is they are not dependent on any variable.

Answer:

Option C.  failing to reject a false null hypothesis

Step-by-step explanation:

Type II error states that Probability of accepting null hypothesis given the fact that null hypothesis is false.

This is considered to be the most important error.

So, from the option given to us option C matches that failing to reject a false null hypothesis.

Answer:

The probability that between 79% and 81%of the 500 sampled wildlife watchers actively observed mammals in 2011 is P=0.412.

Step-by-step explanation:

We know the population proportion π=0.8, according to the 2011 National Survey of Fishing, Hunting, and Wildlife-Associated Recreation.

If we take a sample from this population, and assuming the proportion is correct, it is expected that the sample's proportion to be equal to the population's proportion.

The standard deviation of the sample is equal to:

[tex]\sigma_s=\sqrt{\frac{\pi(1-\pi)}{N}}=\sqrt{\frac{0.8*0.2}{500}}=0.018[/tex]

With the mean and the standard deviaion of the sample, we can calculate the z-value for 0.79 and 0.81:

[tex]z=\frac{p-\pi}{\sigma}=\frac{0.79-0.80}{0.018} =\frac{-0.01}{0.018} = -0.55\\\\z=\frac{p-\pi}{\sigma}=\frac{0.81-0.80}{0.018} =\frac{0.01}{0.018} = 0.55[/tex]

Then, the probability that between 79% and 81%of the 500 sampled wildlife watchers actively observed mammals in 2011 is:

[tex]P(0.79<p<0.81)=P(-0.55<z<0.55)\\\\P(0.79<p<0.81)=P(z<0.55)-P(z<-0.55)\\\\P(0.79<p<0.81)=0.70884-0.29116=0.41768[/tex]

Following are the solution to the given question:

By using the approximation of normal for the proportions so, the values:  

[tex]\to P(\hat{p} < p ) = P(Z < \frac{(\hat{p} - p)}{\sqrt{\frac{p ( 1 - p)}{n}}} )[/tex]

So,

[tex]\to P(0.79 < \hat{p} < 0.81) = P( \hat{p} < 0.81) - P( \hat{p} < 0.79)[/tex]

                                  [tex]= P(Z<\frac{( 0.81 - 0.80)}{\sqrt{\frac{0.80( 1 - 0.80)}{500}}}) - P(Z < \frac{( 0.79 - 0.80)}{ \sqrt{ \frac{0.80 ( 1 - 0.80)}{500}}}) \\\\[/tex]

                                   [tex]= P(Z < 0.56) - P(Z < -0.56)\\\\= 0.7123 - 0.2877 \ \ \text{(using the Z table)}\\\\= 0.4246[/tex]

Therefore, the final answer is "0.4246".

Learn more:

brainly.com/question/3193720

Answer:

We are interested on this probability

[tex]P(X<\mu -2\sigma = 36.6 -2*1.2 =34.2)[/tex]

And the best way to solve this problem is using the normal standard distribution and the z score given by:

[tex]z=\frac{x-\mu}{\sigma}[/tex]

If we apply this formula to our probability we got this:

[tex]P(X<34.2)=P(\frac{X-\mu}{\sigma}<\frac{34.2-\mu}{\sigma})=P(Z<\frac{34.2-36.6}{1.2})=P(z<-2)[/tex]

And we can find this probability using the normal standard table or excel and we got:

[tex]P(z<-2)=0.025[/tex]

And the best answer for this case would be:

2.5%

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 head circumference of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(36.6,1.2)[/tex]  

Where [tex]\mu=36.6[/tex] and [tex]\sigma=1.2[/tex]

We are interested on this probability

[tex]P(X<\mu -2\sigma = 36.6 -2*1.2 =34.2)[/tex]

And the best way to solve this problem is using the normal standard distribution and the z score given by:

[tex]z=\frac{x-\mu}{\sigma}[/tex]

If we apply this formula to our probability we got this:

[tex]P(X<34.2)=P(\frac{X-\mu}{\sigma}<\frac{34.2-\mu}{\sigma})=P(Z<\frac{34.2-36.6}{1.2})=P(z<-2)[/tex]

And we can find this probability using the normal standard table or excel and we got:

[tex]P(z<-2)=0.025[/tex]

And the best answer for this case would be:

2.5%

Answer:

lateral​ surface area of the cone =49π

Step-by-step explanation:

y= x/4

dx/dy = 4

Given x range as  from 0 ≤ x ≤ 7 ⇒ x ranges between (a,b)

a= x/4 = 0/4 = 0

b = x/4= 7/4

lateral​ surface area of the cone

[tex]\int\limits^b_a {2\pi x\sqrt{1 + (\frac{dx}{dy})^2 } } \, dx \\\\= \int\limits^{\frac{7}{4}}_0 {2\pi (4y)\sqrt{1 + (4)^2 } } \, dx \\\\= 32\pi \int\limits^{\frac{7}{4}}_0 { y } \, dx \\\\= 32\pi [\frac{y^2}{2}]|^\frac{7}{4}_0[/tex]

=32π * 49/(16 *2)

=49π

Final answer:

To find the lateral surface area of the cone with the given line segment, use the formula [tex]L = \(\ r l\)[/tex], where r is the base radius, calculated from the given equation, and l is the slant height determined by the Pythagorean theorem.

Explanation:

The student has asked for help in finding the lateral surface area of a cone generated by revolving a line segment around the x-axis.

The specific line segment is given by the equation [tex]y = \frac{1}{4} x[/tex], for 0 \<= x \<= 7. To find this surface area, we can use the formulas for the surface area of a cone, which is \rl\, where \ is the radius of the base of the cone and \ is the slant height.

In this case, when x=7, y = [tex]y = \frac{1}{4} x[/tex] = 7/4, which gives us the radius of the base of the cone. The slant height can be calculated using the Pythagorean theorem, since we have a right triangle with the slant height as the hypotenuse, and the radius and height of the cone as the other two sides. Thus, [tex]l = {7^2+(7/4)^2}[/tex]

Finally, the formula for the lateral surface area of the cone is L = rl, which, after substituting the given values, results in the exact answer needed. The calculation will result in the following:

[tex]L = \frac{7}{4}\times\sqrt{7^2+(7/4)^2}[/tex]

Answer:

(a) The value of P (X = 7) is 0.1388.

(b) The value of P (X ≥ 3) is 0.9380.

(c) The value of P (2 < X < 7) is 0.5433.

(d) [tex]\mu_{X}=6[/tex]

(e) [tex]\sigma_{X}=2.45[/tex]

Step-by-step explanation:

Let X = number of uranium fission tracks on per cm² surface area of the mineral.

The average number of track per cm² surface area is, λ = 6.

The random variable X follows a Poisson distribution with parameter λ = 6.

The probability mass function of a Poisson distribution is:

[tex]P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!};\ x=0, 1, 2, 3...[/tex]

(a)

Compute the value of P (X = 7) as follows:

[tex]P(X=6)=\frac{e^{-6}(6)^{7}}{7!}=\frac{0.0025\times 279936}{5040}=0.1388[/tex]

Thus, the value of P (X = 7) is 0.1388.

(b)

Compute the value of P (X ≥ 3) as follows:

P (X ≥ 3) = 1 - P (X < 3)

              = 1 - P (X = 0) - P (X = 1) - P (X = 2)

              [tex]=1-\frac{e^{-6}(6)^{0}}{0!}-\frac{e^{-6}(6)^{1}}{1!}-\frac{e^{-6}(6)^{2}}{2!}\\=1-0.00248-0.01487-0.04462\\=0.93803\\\approx0.9380[/tex]

Thus, the value of P (X ≥ 3) is 0.9380.

(c)

Compute the value of P (2 < X < 7) as follows:

P (2 < X < 7) = P (X = 3) + P (X = 4) + P (X = 5) + P (X = 6)

                   [tex]=\frac{e^{-6}(6)^{3}}{3!}+\frac{e^{-6}(6)^{4}}{4!}+\frac{e^{-6}(6)^{5}}{5!}+\frac{e^{-6}(6)^{6}}{6!}\\=0.08924+0.13385+0.16062+0.16062\\=0.54433\\\approx0.5443[/tex]

Thus, the value of P (2 < X < 7) is 0.5433.

(d)

The mean of the Poisson distribution is:

[tex]\mu_{X}=\lambda=6[/tex]

(e)

The standard deviation of the Poisson distribution is:

[tex]\sigma_{X}=\sqrt{\sigma^{2}_{X}}=\sqrt{\lambda}=\sqrt{6}=2.4495\approx2.45[/tex]