Answer:
there is no stagnation point
Step-by-step explanation:
for the velocity field V→(u,v)= (0.46 + 2.1x)i→ + (−2.8 - 2.1y)j , the stagnation point is found when the velocity vectors converge in one point ( thus also stays in that place when the point is reached). Thus the stagnation point can be found when the divergence of the velocity field is <0 ( thus it does not diverge , but converges)
div(V) = ∇*V= d/dx (0.46 + 2.1x) + d/dy (−2.8 - 2.1y) = 0
2.1 - 2.1 = 0
since div(V) can never be <0 , there is no stagnation point
Suppose that 20% of the adult women in the United States dye or highlight their hair. We would like to know the probability that a SRS of size 200 would come within plus or minus 3 percentage points of this true value. In other words, find probability that pˆ takes a value between 0.17 and 0.23.
Answer:
71.08% probability that pˆ takes a value between 0.17 and 0.23.
Step-by-step explanation:
We use the binomial approxiation 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]p = 0.2, n = 200[/tex]. So
[tex]\mu = E(X) = np = 200*0.2 = 40[/tex]
[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{200*0.2*0.8} = 5.66[/tex]
In other words, find probability that pˆ takes a value between 0.17 and 0.23.
This probability is the pvalue of Z when X = 200*0.23 = 46 subtracted by the pvalue of Z when X = 200*0.17 = 34. So
X = 46
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{46 - 40}{5.66}[/tex]
[tex]Z = 1.06[/tex]
[tex]Z = 1.06[/tex] has a pvalue of 0.8554
X = 34
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{34 - 40}{5.66}[/tex]
[tex]Z = -1.06[/tex]
[tex]Z = -1.06[/tex] has a pvalue of 0.1446
0.8554 - 0.1446 = 0.7108
71.08% probability that pˆ takes a value between 0.17 and 0.23.
The probability that [tex]\( \hat{p} \)[/tex] takes a value between 0.17 and 0.23 is approximately 0.7108.
1. Given that 20% of adult women dye or highlight their hair, the true population proportion [tex]\( p \)[/tex] is 0.20.
2. We want to find the probability that a sample proportion [tex]\( \hat{p} \)[/tex] from a simple random sample (SRS) of size 200 falls within plus or minus 3 percentage points of this true value. In other words, we want to find [tex]\( P(0.17 < \hat{p} < 0.23) \)[/tex].
3. The standard error of [tex]\( \hat{p} \)[/tex] is given by:
[tex]\[ SE = \sqrt{\frac{p(1-p)}{n}} \][/tex]
where [tex]\( p = 0.20 \)[/tex] (the true population proportion) and [tex]\( n = 200 \)[/tex] (the sample size).
4. Calculate the standard error:
[tex]\[ SE = \sqrt{\frac{0.20 \times (1-0.20)}{200}} = \sqrt{\frac{0.20 \times 0.80}{200}} \][/tex]
[tex]\[ SE = \sqrt{\frac{0.16}{200}} = \sqrt{0.0008} \approx 0.0283 \][/tex]
5. Next, we find the z-scores corresponding to the values 0.17 and 0.23 using the standard normal distribution table:
[tex]\[ z_{0.17} = \frac{0.17 - 0.20}{0.0283} \approx -1.0601 \][/tex]
[tex]\[ z_{0.23} = \frac{0.23 - 0.20}{0.0283} \approx 1.0601 \][/tex]
6. Using the z-scores, we find the corresponding probabilities from the standard normal distribution table:
[tex]\[ P(\hat{p} < 0.17) \approx P(Z < -1.0601) \approx 0.1446 \][/tex]
[tex]\[ P(\hat{p} < 0.23) \approx P(Z < 1.0601) \approx 0.8554 \][/tex]
7. Therefore, the probability that [tex]\( \hat{p} \)[/tex] takes a value between 0.17 and 0.23 is approximately:
[tex]\[ P(0.17 < \hat{p} < 0.23) = P(\hat{p} < 0.23) - P(\hat{p} < 0.17) \][/tex]
[tex]\[ \approx 0.8554 - 0.1446 = 0.7108 \][/tex]
8. Alternatively, we can find this probability directly using the cumulative distribution function (CDF) of the standard normal distribution:
[tex]\[ P(0.17 < \hat{p} < 0.23) = P(-1.0601 < Z < 1.0601) \][/tex]
[tex]\[ \approx \Phi(1.0601) - \Phi(-1.0601) \][/tex]
[tex]\[ \approx 0.8554 - 0.1446 = 0.7108 \][/tex]
9. Therefore, the probability that[tex]\( \hat{p} \)[/tex] takes a value between 0.17 and 0.23 is approximately 0.7108.
Consumer complaints are frequently reported to the Better Business Bureau. Industries with the most complaints to the Better Business Bureau are often banks, cable and satellite television companies, collection agencies, cellular phone providers, and new car dealerships. The results for a sample of 200 complaints are in the file BBB.
(a) Show the frequency and percent frequency of complaints by industry.
Industry Frequency % Frequency
Bank
Cable
Car
Cell
Collection
Total
(b) Which industry had the highest number of complaints?
(c) Comment on the percentage frequency distribution for complaints.
Answer:
Step-by-step explanation:
a. The frequency and percent frequency of complaints by industry is shown below:
Industry Frequency Percentage Frequency
Bank 26 13%
Cable 44 22%
Car 42 21%
Cell 60 30%
Collection 48 13%
b. Cell has the highest complaints (60).
c. The decreasing order of complaints goes thus: Cell (60%) , Cable (22%), Car(21%).
Both Collection and Bank, together account nearly up to 27% of the total complaints.
The role of probability in inferential statistics How is probability used in inferential statistics?A researcher uses probability to decide whether the sample she obtained is likely to be a sample from a particular population.A researcher uses probability to decide whether to draw a sample from a population.A researcher uses probability to decide whether to use inferential or descriptive statistics.
Answer:A researcher uses probability to decide whether the sample she obtained is likely to be a sample from a particular population.
Step-by-step explanation: Inferential statistics is a Statistical process used to compare two or more samples or treatments.
Probability helps in inferential statistics to decide whether the sample obtained is likely from the population of interest.
Inferential statistics use data obtained from the sample of interest in a research to compare the treatment or samples.Through Inferential statistics researchers make conclusions about the entire population.
Probability in inferential statistics is used to make inferences about a population based on sample data. It provides the foundation for statistical methods such as confidence intervals and hypothesis testing to evaluate the accuracy of the sample in representing the population.
Probability in inferential statistics is critical in helping researchers make inferences about a population from a sample. When researchers collect data from a sample, they use probability theory to deduce how likely it is that their observations are reflective of the entire population or occured by chance. Inferential statistical methods, such as confidence intervals and hypothesis testing, leverage probability to make these determinations.
Probability enables statisticians to evaluate the accuracy of the sample data in representing the population, decide how confident they can be about their inferences, and test the validity of existing hypotheses about the population parameters based on sample data.
For instance, if an inferential statistical test indicates that the likelihood of obtaining the observed sample results by chance is only 5%, researchers can infer there is a 95% probability that the sample accurately reflects the population, supporting the hypothesis being tested. Therefore, probability is used to determine how much confidence researchers can have in their sample data when making generalizations about a larger group.
A thin sheet of ice is in the form of a circle. If the ice is melting in such a way that the area of the sheet is decreasing at a rate of 0.5 m2/sec at what rate is the radius decreasing when the area of the sheet is 12 m2
Answer:
dx/dt = 0,04 m/sec
Step-by-step explanation:
Area of the circle is:
A(c) =π*x² where x is a radius of the circle
Applying differentiation in relation to time we get:
dA(c)/dt = π*2*x* dx/dt
In this equation we know:
dA(c)/dt = 0,5 m²/sec
And are looking for dx/dt then
0,5 = 2*π*x*dx/dt when the area of the sheet is 12 m² (1)
When A(c) = 12 m² x = ??
A(c) = 12 = π*x² ⇒ 12 = 3.14* x² ⇒ 12/3.14 = x²
x² = 3,82 ⇒ x = √3,82 ⇒ x = 1,954 m
Finally plugging ths value in equation (1)
0,5 = 6,28*1,954*dx/dt
dx/dt = 0,5 /12.28
dx/dt = 0,04 m/sec
The rate at which the radius is decreasing when the area of the sheet is 12 m² is; dr/dt = 0.041 m/s
We are given;
Area of sheet; A = 12 m²
Rate of change of area; dA/dt = 0.5 m²/s
Now, formula for area of the circular sheet is given as;
A = πr²
Thus; 12 = πr²
r = √(12/π)
r = 1.9554 m
Now, we want to find the rate at which the radius is decreasing and so we differentiate both sides of the area formula with respect to t;
dA/dt = 2πr(dr/dt)
Thus;
0.5 = 2π × 1.9554(dr/dt)
dr/dt = 0.5/(2π × 1.9554)
dr/dt = 0.041 m/s
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If one wanted to find the probability of 10 customer arrivals in an hour at a service station, one would generally use the _____. a. hypergeometric probability distribution b. Poisson probability distribution c. exponential probability distribution d. binomial probability distribution
Answer:
b. Poisson probability distribution
Step-by-step explanation:
The Poisson distribution is the discrete probability distribution of the number of events occurring in a given time period. The poisson distribution is also use when few large demand is expected.
In this question; poisson distribution is use to find the probability of 10 customers arrivals in an hour at a service station.
To find the probability of 10 customer arrivals in an hour at a service station, one would use the Poisson probability distribution, which calculates the probability of a certain number of events happening in a set period of time.
Explanation:The correct answer is b. Poisson probability distribution. This type of distribution is often used to calculate the probability of a certain number of events happening in a set period of time. In this case, the 'events' are the arrivals of customers at a service center within an hour.
Here's a very simplified version of the steps to calculate a Poisson probability:
Step 1: Identify the average rate (λ) - this is the average number of times the event is happening per unit of measure (in your case, customer arrivals per hour).
Step 2: Use the formula for Poisson probability, which is P(x; λ) = e^-λ * λ^x / x! Where 'x' is the actual number of successes that result from the experiment, 'e' is approx 2.71828 and '!' denotes a factorial.
So, if we knew the average rate of customer arrivals, we could easily apply it to this formula to get the probability of 10 customer arrivals in an hour.
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Find the area of a regular octagon with side length 8 cm. Round your answer to the nearest square centimeter.
A. 53 cm
B. 106 cm
C. 155 cm
D. 309 cm
Answer:
A≈309.02cm² Step by step
The distribution of the amount of change in UF student's pockets has an average of 2.02 dollars and a standard deviation of 3.00 dollars. Suppose that a random sample of 45 UF students was taken and each was asked to count the change in their pocket. The sampling distribution of the sample mean amount of change in students pockets is
A. approximately normal with a mean of 2.02 dollars and a standard error of 3.00 dollars
B. approximately normal with a mean of 2.02 dollars and a standard error of 0.45 dollars
C. approximately normal with an unknown mean and standard error.
D. not approximately normal
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
Elana owns a consulting business that helps software companies market their services to school districts. She earns an average of $5687.1 for every contract one of her client companies signs with her help. In order to be able to run her business, she needs to cover $7,000/month (rent, licenses, etc.). The average costs associated with each contract are $1260.7. How many contracts must she facilitate each month in order to break even
Answer:
2 contracts
Step-by-step explanation:
Break even point refers to the number of units or sales that needs to be generated for the company to make neither a profit nor a loss.
This means that at the break even point, sales is equivalent to the cost incurred (both fixed and variable).
Let the number of contracts that must be signed to break even be s
The rent is a fixed cost while the cost associated with each contract is variable.
5687.1s = 7000 + 1260.7s
5687.1s - 1260.7s = 7000
4426.4 s = 7000
s = 1.58
≈ 2
She must facilitate 2 contracts each month to break even.
The error in the length of a part (absolute value of the difference between the actual length and the target length), in mm, is a random variable with probability density function 120x2-x 0 〈 x 〈 1 f(x) = 0 otherwise a. What is the probability that the error is less than 0.2 mm? b. Find the mean error. c. Find the variance of the error. d. Find the cumulative distribution function of the error. e.The specification for the error is 0 to 0.8 mm, What is the probability that the specification is met?
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.
A retail store stocks windbreaker jackets in small, medium, large, and extra large and all are available in blue or red. What are the combined choices and how many combined choices are there?
Answer:
Choices={SB,SR,MB,MR,LS,LM,XLB,XLR}
8 combined choices
Step-by-step explanation:
Combinations
We'll define two sets of options for the windbreaker jackets, one for the sizes and another for the colors. Being S=small, M=medium, L=large, and XL=extra large, then
Z={S,M,L,XL}
is the set of possible sizes for the windbreaker jackets. Now, being B=blue and R=red, the set of colors is
C={B,R}
The combined choices are found by the cartesian product of ZxC:
Choices={SB,SR,MB,MR,LS,LM,XLB,XLR}
Where MB, for example, is Medium-Blue
That is the sample space for all the possible combinations, there are 8 in total
According to the Counting Principle in Mathematics, there are 8 different combinations of sizes and colors for the windbreaker jackets available at the retail store.
Explanation:The retail store offers windbreaker jackets in four different sizes: small, medium, large, and extra large. Each of these sizes is available in two colors: blue and red. Therefore, using a concept in mathematics known as the Counting Principle, we can ascertain the number of combinations. The Counting Principle states that if one event can occur in m ways and another can occur in n ways, then the number of ways that both events can occur is m*n.
So in this scenario, we have 4 sizes (small, medium, large, extra-large) and 2 colors (blue, red). Applying the Counting Principle, there are a total of 4 * 2 = 8 different combinations of jackets that can be purchased from the store.
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a) What percentage of the area under the normal curve lies to the left of μ? % (b) What percentage of the area under the normal curve lies between μ − σ and μ + σ? % (c) What percentage of the area under the normal curve lies between μ − 3σ and μ + 3σ? %
Answer:
a) 50%
b) 68%
c) 99%
Step-by-step explanation:
for a standard normal curve ,
a) since the standard normal curve is symmetric and centred around μ , 50% of the curve lies at the left of μ and 50% lies to the right
b) according to the 68-95-99 rule, 68% of the standard normal curve lies from μ − σ and μ + σ
c) from the same rule , 99% of the standard normal curve lies from μ − 3σ and μ + 3σ
Answer:
a) 50%
b) 68%
c) 99.7%
Step-by-step explanation:
The Empirical Rule states that, for a normally distributed 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.
The normal distribution is also symmetric, which means that 50% of the measures are below the mean and 50% are above.
In this problem, we have that:
Mean μ
Standard deviation σ
Area under the normal curve = percentage
a) What percentage of the area under the normal curve lies to the left of μ?
Normal distribution is symmetric, so the answer is 50%.
(b) What percentage of the area under the normal curve lies between μ − σ and μ + σ?
Within 1 standard deviation of the mean, so 68%.
(c) What percentage of the area under the normal curve lies between μ − 3σ and μ + 3σ?
Within 3 standard deviation of the mean, so 99.7%.
subtract and simplify. please help
Answer:4y^2-10y-1
Step-by-step explanation:
Give as good a big-O estimate as possible for each of these functions. For the function g in your estimate f(x) ∈ O(g(x)), the function should be much simpler than f. (a) f(n) = (n log n + n 2 )(n 3 + 2)
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.
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An engineer commutes daily from her suburban home to her midtown office. The average time for a one-way trip is 36 minutes, with a standard deviation of 4.9 minutes. Assume the distribution of trip times to be normally distributed. What is the probability that a trip will take at least 35 minutes
Answer:
57.93% probability that a trip will take at least 35 minutes.
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 = 36, \sigma = 4.9[/tex]
What is the probability that a trip will take at least 35 minutes
This probability is 1 subtracted by the pvalue of Z when X = 35. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{35 - 36}{4.9}[/tex]
[tex]Z = -0.2[/tex]
[tex]Z = -0.2[/tex] has a pvalue of 0.4207
1 - 0.4207 = 0.5793
57.93% probability that a trip will take at least 35 minutes.
A type of long-range radio transmits data across the Atlantic Ocean. The number of errors in the transmission during any given amount of time approximately follows a Poisson distribution. The mean number of errors is 2 per hour. (a) What is the probability of having at least 3
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%.
Morgan Bowie is trying to determine the amount to set aside so that he will have enough money on hand in 3 years to overhaul the engine on his vintage used car. While there is some uncertainty about the cost of engine overhauls in 3 years, by conducting some research online, Morgan has developed the following estimates.
Engine Overhaul Estimated Cash Outflow Probability Assessment
$390 10%
570 30%
750 50%
790 10%
How much should Keith Bowie deposit today in an account earning 6%, compounded annually, so that he will have enough money on hand in 6 years to pay for the overhaul?
Final answer:
Keith Bowie should deposit $667.45 into an account today to have enough money in 6 years to pay for the engine overhaul.
Explanation:
To determine the amount Keith Bowie should deposit today in an account earning 6% interest compounded annually, we can use the concept of present value. The present value of a future cash flow is calculated by dividing the future cash flow by (1 + interest rate) raised to the power of the number of years. In this case, we want to find the amount needed that would accumulate to cover the engine overhaul cost in 6 years. Let's assume the cost of the overhaul is $390 with a 10% probability, $570 with a 30% probability, $750 with a 50% probability, and $790 with a 10% probability. We can calculate the present value by multiplying each cash flow amount with its respective probability, discounting it back at the interest rate over 6 years, and adding them all up:
Present Value = (390 * 0.10) / (1 + 0.06)6 + (570 * 0.30) / (1 + 0.06)6 + (750 * 0.50) / (1 + 0.06)6 + (790 * 0.10) / (1 + 0.06)6 = $667.45
Therefore, Keith Bowie should deposit $667.45 into the account today to ensure he has enough money on hand in 6 years to pay for the engine overhaul.
The cost, in dollars, of producing x yards of a certain fabric is C(x) = 1500 + 12x − 0.1x2 + 0.0005x3. (a) Find the marginal cost function. C'(x) = (b) Find C'(500) and explain its meaning. What does it predict? C'(500) = and this is the rate at which costs are increasing with respect to the production level when x = . C'(500) predicts the cost of producing the yard. (c) Compare C'(500) with the cost of manufacturing the 501st yard of fabric. (Round your answers to four decimal places.) The cost of manufacturing the 501st yard of fabric is C(501) − C(500) = − 45,000 ≈ , which is approximately C'(500).
a) The marginal cost function is [tex]\[C'(x) = 12 - 0.2x + 0.0015x^2\][/tex].
b) The value C'(500) = 287 predicts that for every additional yard of fabric produced at this level (500 yards), the cost will increase by $287.
c) The cost of manufacturing the 501st yard of fabric is approximately $45,000, which is approximately equal to the marginal cost C'(500)
a)
To find the marginal cost function, we need to take the derivative of the cost function C(x) with respect to x:
[tex]\[C(x) = 1500 + 12x - 0.1x^2 + 0.0005x^3\][/tex]
Taking the derivative:
[tex]\[C'(x) = \frac{d}{dx}(1500 + 12x - 0.1x^2 + 0.0005x^3)\][/tex]
[tex]\[C'(x) = 12 - 0.2x + 0.0015x^2\][/tex]
b)
To find [tex]\(C'(500)\)[/tex] substitute x = 500 into the marginal cost function:
[tex]\[C'(500) = 12 - 0.2(500) + 0.0015(500^2)\][/tex]
[tex]\[C'(500) = 12 - 100 + 375\][/tex]
[tex]\[C'(500) = 287\][/tex]
The value C'(500) = 287 represents the rate at which costs are increasing with respect to the production level when x = 500. It predicts that for every additional yard of fabric produced at this level (500 yards), the cost will increase by $287.
c)
To compare C'(500) with the cost of manufacturing the 501st yard of fabric, calculate C(501) - C(500):
[tex]\[C(501) - C(500) = (1500 + 12(501) - 0.1(501)^2 + 0.0005(501)^3) - (1500 + 12(500) - 0.1(500)^2 + 0.0005(500)^3)\][/tex]
Calculate the difference:
[tex]\[C(501) - C(500) = -45,000\][/tex]
Therefore, the cost of manufacturing the 501st yard of fabric is approximately $45,000, which is approximately equal to the marginal cost C'(500) when rounded to four decimal places.
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(a) Marginal cost function: C'(x) = 12 - 0.2x + 0.0015x².
(b) C'(500) predicts a $287 cost increase for the 500th yard.
(c) Cost for the 501st yard is approximately $512, slightly higher than C'(500).
(a) To find the marginal cost function, we need to calculate the derivative of the cost function C(x) with respect to x:
C(x) = 1500 + 12x - 0.1x² + 0.0005x³
C'(x) = d/dx [1500 + 12x - 0.1x² + 0.0005x³]
C'(x) = 12 - 0.2x + 0.0015x²
So, the marginal cost function is C'(x) = 12 - 0.2x + 0.0015x².
(b) To find C'(500), we plug in x = 500 into the marginal cost function:
C'(500) = 12 - 0.2(500) + 0.0015(500)²
C'(500) = 12 - 100 + 375
C'(500) = 287
C'(500) represents the rate at which costs are increasing with respect to the production level when 500 yards of fabric are produced. In this case, it predicts that the cost is increasing at a rate of $287 per yard for the 500th yard produced.
(c) To compare C'(500) with the cost of manufacturing the 501st yard of fabric, we need to calculate C(501) - C(500):
C(501) - C(500) = [1500 + 12(501) - 0.1(501)² + 0.0005(501)³] - [1500 + 12(500) - 0.1(500)² + 0.0005(500)³]
C(501) - C(500) = [1500 + 6012 - 25005 + 627507] - [1500 + 6000 - 25000 + 625000]
C(501) - C(500) = [627012] - [626500]
C(501) - C(500) = 512
So, the cost of manufacturing the 501st yard of fabric is $512, which is approximately equal to C'(500), which was calculated as $287. This means that the cost increased by approximately $287 for the 500th yard and by $512 for the 501st yard.
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4.36 Stats final scores. Each year about 1500 students take the introductory statistics course at a large university. This year scores on the final exam are distributed with a median of 74 points, a mean of 70 points, and a standard deviation of 10 points. There are no students who scored above 100 (the maximum score attainable on the final) but a few students scored below 20 points. (a) Is the distribution of scores on this final exam symmetric, right skewed, or left skewed
Answer:
The distribution of scores on this final exam is left-skewed.
Step-by-step explanation:
We use the Pearson Mode Skewness to solve this question. It states that:
If the median is higher than the mean, the distribution is left-skewed.
If the median is lower than the mean, the distribution is right-skewed.
If the median is the same as the mean, the distribution is symmetric.
In this problem, we have that:
Median = 74
Mean = 70
Median higher than the mean
So the distribution of scores on this final exam is left-skewed.
Answer:
Step-by-step explanation:
median 74
mean 70
this is the answer
Write the composite function in the form f(g(x)).[Identify the inner function u = g(x)and the outer function
y = f(u).]
$ y = e^{{\color{red}7}\sqrt{x}} $
(g(x), f(u)) = ??
and find the derivative
For what values of x does the graph of f have a horizontal tangent? (Use n as your integer variable. Enter your answers as a comma-separated list.)
f(x) = x ? 2 sin x
x=??
Answer:
a) (g(x), f(u)) = ( 7*√x , e^u )
b) y ' = 3.5 * e^(7*√x) / √x
Step-by-step explanation:
Given:
- The given function:
y = e^(7*√x)
Find:
- Express the given function as a composite of f(g(x)). Where, u = g(x) and y = f(u).
- Express the derivative of y, y'?
Solution:
- We will assume the exponent of the natural log to be the u. So u is:
u = g(x) = 7*√x
- Then y is a function of u as follows:
y = f(u) = e^u
- The composite function is as follows:
(g(x), f(u)) = ( 7*√x , e^u )
- The derivative of y is such that:
y = f(g(x))
y' = f' (g(x) ) * g'(x)
y' = f'(u) * g'(x)
y' = e^u* 3.5 / √x
- Hence,
y ' = 3.5 * e^(7*√x) / √x
There are 24 tennis balls in a basket. Four tennis players divided the balls evenly among each other. How many tennis balls did each player get?
A. t - 4 = 24
B. t 4 = 24
C. 4t = 24
D. t + 4 = 24
E. 4 t = 24
F. t = 24 + 4
Answer:
Each player gets 6 tennis balls.
4 × t = 24
Step-by-step explanation:
The total number of tennis balls in the basket is, 24 balls.
The number of tennis players is, 4 players.
On equally dividing all the balls among the 4 players, each player gets,
[tex]No.\ of\ balls\ each\ player\ gets=\frac{No.\ of\ balls}{No.\ of\ players} \\=\frac{24}{4} \\=6[/tex]
Thus, each player gets 6 tennis balls.
If we want to represent the equation of the number of players and number of balls, the equation will be,
4 × t = 24
t = number of tennis balls each player gets.
Final answer:
To divide 24 tennis balls evenly among four players, we use the equation 4t = 24, where 't' is the number of balls per player. By dividing 24 by 4, we find that each player receives 6 tennis balls.
Explanation:
The student is asking how to divide 24 tennis balls evenly among four players, which is a basic division problem in mathematics. The correct mathematical equation to represent this situation is C. 4t = 24, where 't' represents the number of tennis balls each player gets. To find the value of 't', we divide the total number of balls, 24, by the number of players, 4.
We perform the division as follows:
Write down the equation: 4t = 24.
Divide both sides of the equation by 4 to solve for 't': t = 24 / 4.
Calculate the result: t = 6.
Each player gets 6 tennis balls.
Blood pressure: High blood pressure has been identified as a risk factor for heart attacks and strokes. The proportion of U.S. adults with high blood pressure is 0.2. A sample of 37 U.S. adults is chosen. Use the TI-84 Plus Calculator as needed. Round the answer to at least four decimal places.
Answer:
[tex]p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex]
For this case we know this:
[tex] n=37 , p=0.2[/tex]
We can find the standard error like this:
[tex] SE = \sqrt{\frac{\hat p (1-\hat p)}{n}}= \sqrt{\frac{0.2*0.8}{37}}= 0.0658[/tex]
So then our random variable can be described as:
[tex] p \sim N(0.2, 0.0658)[/tex]
Let's suppose that the question on this case is find the probability that the population proportion would be higher than 0.4:
[tex] P(p>0.4)[/tex]
We can use the z score given by:
[tex] z = \frac{p -\mu_p}{SE_p}[/tex]
And using this we got this:
[tex] P(p>0.4) = 1-P(z< \frac{0.4-0.2}{0.0658}) = 1-P(z<3.04) = 0.0012[/tex]
And we can find this probability using the Ti 84 on this way:
2nd> VARS> DISTR > normalcdf
And the code that we need to use for this case would be:
1-normalcdf(-1000, 3.04; 0;1)
Or equivalently we can use:
1-normalcdf(-1000, 0.4; 0.2;0.0658)
Step-by-step explanation:
We need to check if we can use the normal approximation:
[tex] np = 37 *0.2 = 7.4 \geq 5[/tex]
[tex] n(1-p) = 37*0.8 = 29.6\geq 5[/tex]
We assume independence on each event and a random sampling method so we can conclude that we can use the normal approximation and then ,the population proportion have the following distribution :
[tex]p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex]
For this case we know this:
[tex] n=37 , p=0.2[/tex]
We can find the standard error like this:
[tex] SE = \sqrt{\frac{\hat p (1-\hat p)}{n}}= \sqrt{\frac{0.2*0.8}{37}}= 0.0658[/tex]
So then our random variable can be described as:
[tex] p \sim N(0.2, 0.0658)[/tex]
Let's suppose that the question on this case is find the probability that the population proportion would be higher than 0.4:
[tex] P(p>0.4)[/tex]
We can use the z score given by:
[tex] z = \frac{p -\mu_p}{SE_p}[/tex]
And using this we got this:
[tex] P(p>0.4) = 1-P(z< \frac{0.4-0.2}{0.0658}) = 1-P(z<3.04) = 0.0012[/tex]
And we can find this probability using the Ti 84 on this way:
2nd> VARS> DISTR > normalcdf
And the code that we need to use for this case would be:
1-normalcdf(-1000, 3.04; 0;1)
Or equivalently we can use:
1-normalcdf(-1000, 0.4; 0.2;0.0658)
The following data represent the muzzle velocity (in feet per second) of shells fired from a 155-mm gun. For each shell, two measurements of the velocity were recorded using two different measuring devices, resulting in the following data.
Observation 1 2 3 4 5 6
A 792.5 793.8 792.5 791.9 792.2 791.1
B 795.8 797.1 793.6 797.3 789.9 797.9
(a) Construct a 99% confidence interval about the population mean difference. Compute the difference as device A minus device B. Interpret your results.
Answer:
note:
solution is attached in word form due to error in mathematical equation. furthermore i also attach Screenshot of solution in word due to different version of MS Office please find the attachment
Determine the following: A. P(X > 654) for N(650, 10) B. P(Z < 0.72) Solution: A. 1 – pnorm(654,650,10) = 0.3445783 B. qnorm(0.72) = 0.5828415 What was done wrong in this solution?
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,
According to the 2011 National Survey of Fishing, Hunting, and Wildlife-Associated Recreation, there were over 71 million wildife watchers in the US. Of these wildlife watchers, the survey reports that 80% actively observed mammals. Suppose that one of the census workers repeated the survey with a simple random sample of only 500 wildlife watchers that same year. Assuming that the original survey's 80% claim is correct, what is the approximate probability that between 79% and 81%of the 500 sampled wildlife watchers actively observed mammals in 2011?
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".
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Which polynomials are in standard form?
Choose all answers that apply:
5-2x
x^4-8x^2 -16
5x^3 + 4x^4 — 3x + 1
None of the above
Answer: The second option, x^4 - 8x^2 - 16.
Step-by-step explanation:
Polynomials in standard form start with the highest degree, from greatest to least exponent. After all terms with exponents are in order, alphabetical variables are next. In this case there's only x. Last are constant terms, which are by itself, with no variable next to it/an exponent to the right of it.
x^4 - 8x^2 - 16 is in standard form because it follows the criteria above. 4 is the highest degree since it's the highest exponent in the polynomial expression, which is why it starts off with x^4. Other terms with lesser exponents are next. In this case, it's 8x^2 with the less exponent of 2. Finally, it ends with your constant term, -16.
The standard form of the polynomial is,
⇒ x⁴ - 8x² - 16
Given that,
All the polynomials are,
⇒ 5 - 2x
⇒ x⁴ - 8x² - 16
⇒ 5x³ + 4x⁴ - 3x + 1
Since we know that,
In standard form, a polynomial is arranged in descending order of the exponents of its terms.
This means that the term with the highest degree is listed first, followed by the terms with lower degrees.
Hence, Based on this definition, the polynomial in standard form among the options is,
⇒ x⁴ - 8x² - 16
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The equation r(t)= (3t+9)i+(sqrt(2)t)j+(t^2)k is the position of a particle in space at time t. Find the angle between the velocity and acceleration vectors at time t=0
What is the angle? ______radians
Answer:
[tex]\theta= \frac{\pi}{2} +\pi \cdot i[/tex], for all [tex]i = \mathbb{Z} \cup\{0\}[/tex]
Step-by-step explanation:
The velocity vector is found by deriving the position vector depending on the time:
[tex]\dot r(t)= v (t) = 3 \cdot i +\sqrt{2} \cdot j + 2\cdot t \cdot k[/tex]
In turn, acceleration vector is found by deriving the velocity vector depending on time:
[tex]\ddot r(t) = \dot v(t) = a(t) = 2 \cdot k[/tex]
Velocity and acceleration vectors at [tex]t = 0[/tex] are:
[tex]v(0) = 3\cdot i + \sqrt{2} \cdot j\\a(0) = 2 \cdot k\\[/tex]
Norms of both vectors are, respectively:
[tex]||v(0)||\approx 3.317\\||a(0)|| \approx 2[/tex]
The angle between both vectors is determined by using the following characteristic of a Dot Product:
[tex]\theta = \cos^{-1}(\frac{v(0) \bullet a(0)}{||v(0)||\cdot ||a(0)||})[/tex]
Given that cosine has a periodicity of [tex]\pi[/tex]. There is a family of solutions with the form:
[tex]\theta= \frac{\pi}{2} +\pi \cdot i[/tex], for all [tex]i = \mathbb{Z} \cup\{0\}[/tex]
Final answer:
π/2 radians, indicating the vectors are perpendicular at that instant.
Explanation:
The angle between the velocity and acceleration vectors at a given time can be found by first determining the velocity (νt) and acceleration (ν2t) vectors as the first and second derivatives of the position vector r(t). At time t=0, these derivatives can be calculated and then used to find the angle through the dot product and magnitude of these vectors.
For the given position vector
r(t) = (3t+9)i + (√2t)j + (t2)k,
the velocity vector v(t) is obtained by differentiating each component of r(t) with respect to time t, which gives
v(t) = (3)i + (√2)j + (2t)k.
Similarly, acceleration a(t) is the derivative of velocity v(t), which results in a(t) = (0)i + (0)j + (2)k.
At t=0, v(0) = (3)i + (√2)j and a(0) = (2)k. The angle θ between v(0) and a(0) is given by the cosine of the angle between the two vectors, which is calculated using the dot product formula:
θ = cos-1((v ⋅ a) / (|v||a|)).
Here, (v ⋅ a) is the dot product of v(0) and a(0), and |v| and |a| are the magnitudes of v(0) and a(0), respectively.
Since v(0) and a(0) are perpendicular at t=0, their dot product is 0, and the magnitudes of v(0) and a(0) do not affect the angle. Therefore, the angle θ is simply cos-1(0), which is π/2 radians, indicating the vectors are perpendicular.
4.47 Consider an experiment, the events A and B, and probabilities P(A) 5 0.55, P(B) 5 0.45, and P(A d B) 5 0.15. Find the probability of: a. A or B occurring. b. A and B occurring. c. Just A occurring. d. Just A or just B occurring.
Answer:
a) 0.85
b) 0.15
c) 0.40
d) 0.70
Step-by-step explanation:
P(A) = 0.55
P(B) = 0.45
P(A n B) = 0.15
a) Probability of A or B occurring = P(A u B) = P(A) + P(B) - P(A n B) = 0.55 + 0.45 - 0.15 = 0.85
b) Probability of A and B occurring = P(A n B) = 0.15
c) Probability of just A occurring = P(A n B') = P(A) - P(A n B) = 0.55 - 0.15 = 0.40
d) Probability of just A or just B occurring = P(A n B') + P(A' n B) = 0.4 + (0.45 - 0.15) = 0.4 + 0.3 = 0.70
There are two sources that serve as the foundation for conducting research on learning. The first source addresses characteristics of knowledge itself and the different ways in which we learn things. The second source focuses on what goes on in our minds and how that is theoretically represented.
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.
Explanation: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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Seventy percent of light aircraft that disappear while in flight in a certain country are subsequently discovered. Of the aircraft that are discovered, 60% have an emergency locator, whereas 90% of the aircraft not discovered do not have an emergency locator. Suppose that a light aircraft has disappeared.a) If it has an emergency locator, what is the probability that it will not be discovered?b) If it does not have an emergency locator, what is the probability that it will be discovered?c) If we consider 10 light aircraft that disappeared in flight with an emergency recorder, what is the probability that 7 of them are discovered?
Answer:
Figure out the various probabilities first, that will make the rest of the questions easier:
P(discovered) = .7
P(not discovered) = 1 - .7 = .3
P(locator|discovered) = .6
P(no locator|discovered) = 1 - .6 = .4
P(locator|not discovered) = 1 - .9 = .1
P(no locator|not discovered) = .9
P(discovered and locator) = .7 * .6 = .42
P(discovered and no locator) = .7 * .4 = .28
P(not discovered and locator) = .3 * .1 = .03
P(not discovered and no locator) = .3 * .9 = .27
a) The total probability that an aircraft has a locator is .42 + .03 = .45. So the probability it will not be discovered, given it has a locator, is .03/.45 = .067
b) The total probability that an aircraft does not have a locator is .28 + .27 = .55. So the probability it will be discovered, given it does not have a locator, is .28/.55 = .509
c) Probability that 7 are discovered = C(10,7) * P(discovered|locator)^7 * P(not discovered|locator)^3
We already figured out P(not discovered|locator) = .067, so P(discovered|locator) = 1-.067 = .933. C(10,7) = 10*9*8, so we can compute total probability: 10*9*8 * .933^7 * .067^3 = .133
Step-by-step explanation:
Birth weights of full-term babies in a certain area are normally distributed with mean 7.13 pounds and standard deviation 1.29 pounds. A newborn weighing 5.5 pounds or less is a low-weight baby. What is the probability that a randomly selected newborn is low-weight? Use the appropriate applet. Enter a number in decimal form, e.g. 0.68, not 68 or 68%.
Answer: probability that a randomly selected newborn is low-weight is 0.1038
Step-by-step explanation:
Since Birth weights of full-term babies in a certain area are normally distributed m, we would apply the formula for normal distribution which is expressed as
z = (x - µ)/σ
Where
x = birth weights of full-term babies.
µ = mean weight
σ = standard deviation
From the information given,
µ = 7.13 pounds
σ = 1.29 pounds
The probability that a randomly selected newborn is low-weight is expressed as
P(x ≤ 5.5)
For x = 5.5
z = (5.5 - 7.13)/1.29 = - 1.26
Looking at the normal distribution table, the probability corresponding to the z score is 0.1038
P(x ≤ 5.5) = 0.1038