The number of accidents per week at a hazardous intersection varies with mean 2.2 and standard deviation 1.1. This distribution takes only whole-number values, so it is certainly not Normal. (a) Let x be the mean number of accidents per week at the intersection during a year (52 weeks). What is the approximate distribution of x according to the central limit theorem

Answer: 63 defective widgets

Step-by-step explanation:

Given that the proportion should not exceed 5%, that is:

p< or = 5%.

So we take p = 5% = 0.05

q = 1 - 0.05 = 0.095

Where q is the proportion of non-defective

We need to calculate the standard error (standard deviation)

S = √pq/n

Where n = 1024

S = √(0.05 × 0.095)/1024

S = 0.00681

Since production is to maximize profit(profitable), we need to minimize the number of defective items. So we find the limit of defective product to make this possible using the Upper Class Limit.

UCL = p + Za/2(n-1) × S

Where a is alpha of confidence interval = 100 -90 = 10%

a/2 = 5% = 0.05

UCL = p + Z (0.05) × 0.00681

Z(0.05) is read on the t-distribution table at (n-1) degree of freedom, which is at infinity since 1023 = n-1 is large.

Z a/2 = 1.64

UCL = 0.05 + 1.64 × 0.00681

UCL = 0.0612

Since the UCL in this case is a measure of proportion of defective widgets

Maximum defective widgets = 0.0612 ×1024 = 63

Alternatively

UCL = p + 3√pq/n

= 0.05 + 3(0.00681)

= 0.05 + 0.02043 = 0.07043

UCL =0.07043

Max. Number of widgets = 0.07043 × 1024

= 72

Answer:

a) the probability is 0.12 (12%)

b) the probability is 0.61 (61%)

c) the probability is 0.476 (47.6%)

a) the probability is 0.538 (53.8%)

Step-by-step explanation:

a) denoting the event H= hits her target and G= a gust of wind appears hen

P(H∩G) = probability that a gust of wind appears * probability of hitting the target given that is windy = 0.3* 0.4 = 0.12 (12%)

b) for any given shot

P(H)= probability that a gust of wind appears*probability of hitting the target given that is windy + probability that a gust of wind does not appear*probability of hitting the target given that is not windy = 0.3*0.4+0.7*0.7 = 0.12+0.49 = 0.61 (61%)

c) denoting P₂  as the probability of hitting once in 2 shots  and since the archer can hit in the first shot or the second , then

P₂ = P(H)*(1-P(H))+ (1-P(H))*P(H) = 2*P(H) *(1-P(H)) = 2*0.61*0.39= 0.476 (47.6%)

d) for conditional probability we can use the theorem of Bayes , where

M= the archer misses the shot → P(M) = 1- P(H) = 0.39

S= it is not windy when the archer shots → P(S) = 1- P(G) = 0.7

then

P(S/M) = P(S∩M)/P(M) = 0.7*(1-0.7)/0.39 = 0.538 (53.8%)

where P(S/M)  is the probability that there was no wind when the archer missed the shot

Answer:

0.0064 is the probability that Isaac will find his first defective can among the first 50 cans.    

Step-by-step explanation:

W are given the following in the question:

Probability of defective can =

[tex]P(A) = \dfrac{1}{90}[/tex]

We have to find the probability that Isaac will find his first defective can among the first 50 cans.

Then the number of adults follows a geometric distribution, where the probability of success on each trial is p, then the probability that the kth trial (out of k trials) is the first success is

[tex]P(X=k) = (1-p)^{k-1}p[/tex]

We have to evaluate:

[tex]P(x = 50)\\= (1-\frac{1}{90})^{50-1}(\frac{1}{90})\\= 0.0064[/tex]

0.0064 is the probability that Isaac will find his first defective can among the first 50 cans.

Answer:

a) 0.9772 = 97.72% probability that the mean weight of the sample is less than 3.10 pounds

b) 2.88 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], 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 = 3, \sigma = 0.25, n = 25, s = \frac{0.25}{\sqrt{25}} = 0.05[/tex]

a) What is the probability that the mean weight of the sample is less than 3.10 pounds? Round your answer to four decimal places

This is the pvalue of Z when X = 3.10. So

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

By the Central Limit Theorem

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

[tex]Z = \frac{3.1 - 3}{0.05}[/tex]

[tex]Z = 2[/tex]

[tex]Z = 2[/tex] has a pvalue of 0.9772

0.9772 = 97.72% probability that the mean weight of the sample is less than 3.10 pounds

b) What value will the mean weight exceed with probability 0.99? Round your answer to two decimal places.

This is the value of X when Z has a pvalue of 1-0.99 = 0.01. So X when Z = -2.33.

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

[tex]-2.33 = \frac{X - 3}{0.05}[/tex]

[tex]X - 3 = -2.33*0.05[/tex]

[tex]X = 2.88[/tex]

Answer:

1/48 % or 6/48 % chance

Step-by-step explanation:

The probability of winning with the very first Lotto ticket you purchase of the Lotto game consisting of picking 6 of 48 numbers is 1/12271512 or approximately 0.0000000815.

A permutation is a process of calculating the number of ways to choose a set from a larger set in a particular order.

If we want to choose a set of r items from a set of n items in a particular order, we find the permutation nPr = n!/(n-r)!.

A combination is a process of calculating the number of ways to choose a set from a larger set in no particular order.

If we want to choose a set of r items from a set of n items in no particular order, we find the combination nCr = n!/{(r!)(n-r)!}.

In the question, we are asked to determine the probability of winning a lottery by picking 6 numbers from 48 numbers with the first ticket we purchase.

First, we need to calculate the number of combinations of choosing 6 numbers from 48 numbers. As we need to consider no particular order, we will use combinations,

48C6 = 48!/{(6!)(48-6)!} = 48!/(6!*42!) = (43*44*45*46*47*48)/(1*2*3*4*5*6*) (As 48! = 42!*43*44*45*46*47*48, and 42! cancels itself from the numerator and the denominator).

or, 48C6 = 12,271,512.

So, we get the number of combinations = 12,271,512.

We know that we will choose only one particular set of 6 numbers.

∴ The probability of winning on the very first ticket = 1/12,271,512 ≈ 0.0000000815

∴ The probability of winning with the very first Lotto ticket you purchase of the Lotto game consisting of picking 6 of 48 numbers is 1/12271512 or approximately 0.0000000815.

Learn more about the Permutations and Combinations at

https://brainly.com/question/4658834

#SPJ2

Answer:

a. (1.811, 1.929)

Step-by-step explanation:

We have to find find M as such

[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]

In which, z is Z(a/2), [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample. So

[tex]M = 2.58*\frac{0.125}{\sqrt{30}} = 0.059[/tex]

The lower end of the interval is the mean subtracted by M. So it is 1.87 - 0.059 = 1.811.

The upper end of the interval is the mean added to M. So it is 1.87 + 0.059 = 1.929

So the correct answer is:

a. (1.811, 1.929)

Answer:

[tex]E(T)=E(X1 + X2 + X3 + X4 =E(x1) + E(x2) + E(x3) + E(x4) [/tex]

[tex]E(T) =900+900+900+900 =3600[/tex]

And the mean would be:

[tex] \bar X = \frac{T}{4} = \frac{3600}{4}= 900[/tex]

And the standard deviation of total time would be:

[tex]SD(T)=\sqrt(Var(T)) = sqrt(Var(X1) + Var(x2) + Var(x3) + Var(x4))[/tex]

[tex]\sigma=sqrt(30^2+30^2+30^2+30^2) =\sqrt(3600) =60[/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 total life time of t four transistors T = X1 + X2 + X3 + X4 (where X1,X2,X3 and X4 are life time of individual transistors

For this case the mean length of time that four transistors will last

[tex]E(T)=E(X1 + X2 + X3 + X4 =E(x1) + E(x2) + E(x3) + E(x4) [/tex]

[tex]E(T) =900+900+900+900 =3600[/tex]

And the mean would be:

[tex] \bar X = \frac{T}{4} = \frac{3600}{4}= 900[/tex]

And the standard deviation of total time would be:

[tex]SD(T)=\sqrt(Var(T)) = sqrt(Var(X1) + Var(x2) + Var(x3) + Var(x4))[/tex]

[tex]\sigma=sqrt(30^2+30^2+30^2+30^2) =\sqrt(3600) =60[/tex]

The mean lifetime of four transistors is 3600 hours and the standard deviation of the lifetime is 60 hours.

The mean and standard deviation are statistical concepts used to describe a dataset. The mean, or average, is the sum of all data points divided by the number of data points. In this case, the mean lifetime of a single transistor is 900 hours. If you have four transistors, their total lifetime would be four times the mean of a single transistor. So, the mean lifetime for four transistors is 4 * 900 = 3600 hours.

The standard deviation measures the amount of variation in a set of values. When you are considering the lifetime of multiple transistors, you add the variances - not the standard deviations - and then take the square root. The variance is the square of the standard deviation. So, the standard deviation for the lifetime of four transistors is sqrt(4) * 30 = 60 hours.

https://brainly.com/question/33812621

#SPJ11

Answer:

T= A person selected in the poll travel to Europe

NT= A person selected in the poll NOT travel to Europe

For this case we have the following respondents for each event

n(T)= 68

n(NT) = 124

So then the total of people for the poll are:

[tex] n = n(T) + n(NT)= 68 +124= 192[/tex]

And we are interested on the probability of getting someone who has traveled to Europe, and we can use the empirical definition of probability given by:

[tex] p =\frac{Possible}{Total}[/tex]

And if we replace we got:

[tex] p = \frac{n(T)}{n}= \frac{68}{192}= 0.354[/tex]

So then the probability of getting someone who has traveled to Europe is 0.354

Step-by-step explanation:

For this case we define the following events:

T= A person selected in the poll travel to Europe

NT= A person selected in the poll NOT travel to Europe

For this case we have the following respondents for each event

n(T)= 68

n(NT) = 124

So then the total of people for the poll are:

[tex] n = n(T) + n(NT)= 68 +124= 192[/tex]

And we are interested on the probability of getting someone who has traveled to Europe, and we can use the empirical definition of probability given by:

[tex] p =\frac{Possible}{Total}[/tex]

And if we replace we got:

[tex] p = \frac{n(T)}{n}= \frac{68}{192}= 0.354[/tex]

So then the probability of getting someone who has traveled to Europe is 0.354

Answer:

(a) The mean is 52 and the median is 55.

(b) The first quartile is 44 and the third quartile is 60.

(c) The value of range is 33 and the inter-quartile range is 16.

(d) The variance is 100.143 and the standard deviation is 10.01.

(e) There are no outliers in the data set.

(f) Yes

Step-by-step explanation:

The data provided is:

S = {55, 56, 44, 43, 44, 56, 60, 62, 57, 45, 36, 38, 50, 69, 65}

(a)

Compute the mean of the data as follows:

[tex]\bar x=\frac{1}{n}\sum x\\=\frac{1}{15}[55+ 56+ 44+ 43+ 44+ 56+ 60+ 62+ 57+ 45 +36 +38 +50 +69+ 65]\\=\frac{780}{15}\\=52[/tex]

Thus, the mean is 52.

The median for odd set of values is the computed using the formula:

[tex]Median=(\frac{n+1}{2})^{th}\ obs.[/tex]

Arrange the data set in ascending order as follows:

36, 38, 43, 44, 44, 45, 50, 55, 56, 56, 57, 60, 62, 65, 69

There are 15 values in the set.

Compute the median value as follows:

[tex]Median=(\frac{15+1}{2})^{th}\ obs.=(\frac{16}{2})^{th}\ obs.=8^{th}\ observation[/tex]

The 8th observation is, 55.

Thus, the median is 55.

(b)

The first quartile is the middle value of the upper-half of the data set.

The upper-half of the data set is:

36, 38, 43, 44, 44, 45, 50

The middle value of the data set is 44.

Thus, the first quartile is 44.

The third quartile is the middle value of the lower-half of the data set.

The upper-half of the data set is:

56, 56, 57, 60, 62, 65, 69

The middle value of the data set is 60.

Thus, the third quartile is 60.

(c)

The range of a data set is the difference between the maximum and minimum value.

Maximum = 69

Minimum = 36

Compute the value of Range as follows:

[tex]Range =Maximum-Minimum\\=69-36\\=33[/tex]

Thus, the value of range is 33.

The inter-quartile range is the difference between the first and third quartile value.

Compute the value of IQR as follows:

[tex]IQR=Q_{3}-Q_{1}\\=60-44\\=16[/tex]

Thus, the inter-quartile range is 16.

(d)

Compute the variance of the data set as follows:

[tex]s^{2}=\frac{1}{n-1}\sum (x_{i}-\bar x)^{2}\\=\frac{1}{15-1}[(55-52)^{2}+(56-52)^{2}+...+(65-52)^{2}]\\=100.143[/tex]

Thus, the variance is 100.143.

Compute the value of standard deviation as follows:

[tex]s=\sqrt{s^{2}}=\sqrt{100.143}=10.01[/tex]

Thus, the standard deviation is 10.01.

(e)

An outlier is a data value that is different from the remaining values.

An outlier is a value that lies below 1.5 IQR of the first quartile or above 1.5 IQR of the third quartile.

Compute the value of Q₁ - 1.5 IQR as follows:

[tex]Q_{1}-1.5QR=44-1.5\times 16=20[/tex]

Compute the value of Q₃ + 1.5 IQR as follows:

[tex]Q_{3}+1.5QR=60-1.5\times 16=80[/tex]

The minimum value is 36 and the maximum is 69.

None of the values is less than 20 or more than 80.

Thus, there are no outliers in the data set.

(f)

Yes, the data provided indicates that the Walmart is meeting its goal for reducing the number of hourly employees

Answer:

Step-by-step explanation:

Answer:

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

[tex]X \sim N(2.514 cm,0.008 cm)[/tex]  

Where [tex]\mu=2.514[/tex] and [tex]\sigma=0.008[/tex]

b) For this case we can see the interval required on the figure attached,

c) [tex]P(2.49<X<2.51)=P(\frac{2.49-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{2.51-\mu}{\sigma})=P(\frac{2.49-2.514}{0.008}<Z<\frac{2.51-2.514}{0.008})=P(-3<z<-0.5)[/tex]

And we can find this probability with this difference:

[tex]P(-3<z<-0.5)=P(z<-0.5)-P(z<-3)[/tex]

And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.  

[tex]P(-3<z<-0.5)=P(z<-0.5)-P(z<-3)=0.309-0.00135=0.307[/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".  

Part a

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

[tex]X \sim N(2.514 cm,0.008 cm)[/tex]  

Where [tex]\mu=2.514[/tex] and [tex]\sigma=0.008[/tex]

Part b

For this case we can see the interval required on the figure attached,

Part c

We are interested on this probability

[tex]P(2.49<X<2.51)[/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(2.49<X<2.51)=P(\frac{2.49-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{2.51-\mu}{\sigma})=P(\frac{2.49-2.514}{0.008}<Z<\frac{2.51-2.514}{0.008})=P(-3<z<-0.5)[/tex]

And we can find this probability with this difference:

[tex]P(-3<z<-0.5)=P(z<-0.5)-P(z<-3)[/tex]

And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.  

[tex]P(-3<z<-0.5)=P(z<-0.5)-P(z<-3)=0.309-0.00135=0.307[/tex]

Answer:

solution attached below

Step-by-step explanation:

To determine the drag coefficient using the false-position method, start with initial guesses and iterate until the approximate relative error falls below 5%.

To determine the drag coefficient needed for a bungee jumper to have a velocity of 46 m/s after 9 s of free fall using the false-position method, we can follow these steps:

Learn more about Drag coefficient here:

https://brainly.com/question/31824604

#SPJ2

Answer:

(B) 4 minutes after the rocket was launched, the rocket is 516 miles above the ground

Step-by-step explanation:

The function h(t) represents the height of the rocket above the launchpad after a time t minutes.

h(4) = 516 means that when t = 4 minutes, the height h is 516 miles above the launch pad. Note that the time t is measured from when the rocket is launched.

The first option indicates a rate of change, which is a velocity. This is not indicated in the question because velocity is a time derivative of the the height function.

Option C implies that the rocket is 516 miles currently but we do not know what time currently is from the time of launch.

The fourth option has reversed the roles of the variable by implying the time is 516 minutes while the height is 4 miles which is not what the function means.

Answer:

Option a)  H0 : μ = 61 , Hα : μ > 61 .

Step-by-step explanation:

We are given that the time, in number of days, until maturity of a certain variety of tomato plant is Normally distributed, with mean μ and standard deviation s = 2.4.

You select a simple random sample of four plants of this variety and measure the time until maturity. The sample yields x bar = 65.

You read on the package of seeds that these tomatoes reach maturity, on average, in 61 days.

Now, since we know about;

sample size, n = 4

sample standard deviation, s = 2.4

sample, x bar = 65

The only line which represent information about the population mean is reading on the package of seeds that these tomatoes reach maturity, on average, in 61 days.

This shows that population mean , [tex]\mu[/tex] = 61 and in null and alternate hypothesis only population mean is tested.

So, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu[/tex] = 61

Alternate Hypothesis, [tex]H_1[/tex] : [tex]\mu[/tex] > 61 { we are asked for later than expected}

Final answer:

The correct mean time for the first three finishers is 5 minutes, after subtracting the 45-second error from the initially recorded mean time of 5.75 minutes.

Explanation:

The correct mean time for the first three finishers, after adjusting for the stopwatch error, can be calculated by subtracting the 45 seconds from the initially recorded meantime. Since the mean time was calculated to be 5.75 minutes (or 345 seconds), we must correct each time by subtracting 45 seconds and then recalculating the mean.

To find the correct mean time, we do the following steps:

The correct mean time for the first three finishers is therefore 5 minutes.

Answer:

m=1.7

C=68 gr

Step-by-step explanation:

Function Modeling

We are given a relationship between the carbohydrates used by a professional tennis player during a strenuous workout and the time in minutes as 1.7 grams per minute. Being C the carbohydrates in grams and t the time in minutes, the model is

[tex]C=1.7t[/tex]

The slope m of the line is the coefficient of the independent variable, thus m=1.7

The graph of C vs t is shown in the image below.

To find how many carbohydrates the athlete would use in t=40 min, we plug in the value into the equation

[tex]C=1.7\cdot 40=68\ gr[/tex]

The slope of the line representing the rate of carbohydrate usage is 1.7. Multiply this rate (1.7 grams per minute) by the time (40 minutes) to find the total carbohydrates used, which is 68 grams.

The rate at which the tennis player uses carbohydrates is 1.7 grams per minute. In the context of a graph, this rate would represent the slope of the line. So, the slope of the line would be 1.7. Slope, in mathematics, is defined as the change in the y-value (vertical axis) divided by the change in the x-value (horizontal axis). Here, the rate of carbohydrate usage (1.7 grams per minute) is the change in the y-value (carbohydrates used) per change in x-value (time).

Now, you also want to know how many carbohydrates the athlete would use in 40 minutes. We know that the rate of carbohydrate usage is 1.7 grams per minute. So, to find the total amount of carbohydrates used in 40 minutes, you'd simply multiply the rate by the time:

1.7 grams/minute * 40 minutes = 68 grams

So, the athlete would use 68 grams of carbohydrates in 40 minutes.

https://brainly.com/question/31776633

#SPJ3

Answer:

a) 46.1

b) 46

c) Two modes: 46, 41          

Step-by-step explanation:

We are given the following sample of hours per week:

50, 53, 46, 46, 49, 43, 41, 41

a) mean of this data set

[tex]Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}[/tex]

[tex]Mean =\displaystyle\frac{369}{8} = 46.1[/tex]

b) Median of data set

[tex]Median:\\\text{If n is odd, then}\\\\Median = \displaystyle\frac{n+1}{2}th ~term \\\\\text{If n is even, then}\\\\Median = \displaystyle\frac{\frac{n}{2}th~term + (\frac{n}{2}+1)th~term}{2}[/tex]

Sorted data:

41, 41, 43, 46, 46, 49, 50, 53

Median =

[tex]=\dfrac{4^{th}+5^{th}}{2} = \dfrac{46+46}{2} = 46[/tex]

The median of data is 46.

c) Mode of the data set

Mode is the most frequent observation in the data.

The mode of the data are 46 and 41 as they appeared two times.

Thus, there are two modes.

The mean of the data set is 46.1, the median is 46, and there are two modes: 41 and 46.

To answer the questions based on the provided data set of hours worked by eight adults, we need to perform some basic statistical calculations.

(a) Mean

The mean is calculated by summing all the data points and then dividing by the number of data points:

Mean = (50 + 53 + 46 + 46 + 49 + 43 + 41 + 41) / 8 = 369 / 8 = 46.1

(b) Median

First, we need to arrange the data in ascending order: 41, 41, 43, 46, 46, 49, 50, 53. As there are 8 data points (even number), the median is the average of the 4th and 5th values:

Median = (46 + 46) / 2 = 46

(c) Mode

The mode is the number that appears most frequently in the data set. Here, 41 and 46 both appear twice:

There are two modes: 41 and 46 and the mean of the data set is 46.1.

Step-by-step explanation:

Find the area using a proportion.

A / (πr²) = θ / 360°

A / (π (20 ft)²) = 160° / 360°

A ≈ 558.5 ft²

Answer:

Part a

For this case after do the operations we got a value for the correlation coeffcient of:

[tex] r =0.506[/tex]

With this value we can find the determination coefficient:

[tex] r^2 = 0.506^2 = 0.256[/tex]

And with this value we can analyze the proportion of variance explained by one variable and the other. So we can conclude that 25.6% of the mother's BMI variation is explained by the daugther's BMI.

Part a

Since the BMI is a relation between height and weight, other possible variables that can explain the variability are (weight , height, age)  

Step-by-step explanation:

Previous concepts

The correlation coefficient is a "statistical measure that calculates the strength of the relationship between the relative movements of two variables". It's denoted by r and its always between -1 and 1.

And in order to calculate the correlation coefficient we can use this formula:  

[tex]r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]}}[/tex]  

Solution to the problem

Part a

For this case after do the operations we got a value for the correlation coeffcient of:

[tex] r =0.506[/tex]

With this value we can find the determination coefficient:

[tex] r^2 = 0.506^2 = 0.256[/tex]

And with this value we can analyze the proportion of variance explained by one variable and the other. So we can conclude that 25.6% of the mother's BMI variation is explained by the daugther's BMI.

Part a

Since the BMI is a relation between height and weight, other possible variables that can explain the variability are (weight , height, age)  

Answer:

a) 5.48% probability that a baby chosen at random weighs less than 5.5 pounds at birth

b) 0.28% probability that their average birth weight is less than 5.5 pounds

Step-by-step explanation:

To solve this question, the normal probability distribution and the central limit theorem are used.

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 normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sample means 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 = 7.5, \sigma = 1.25[/tex]

(a) What is the probability that a baby chosen at random weighs less than 5.5 pounds at birth?

This is the pvalue of Z when X = 5.5. So

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

[tex]Z = \frac{5.5 - 7.5}{1.25}[/tex]

[tex]Z = -1.6[/tex]

[tex]Z = -1.6[/tex] has a pvalue of 0.0548

5.48% probability that a baby chosen at random weighs less than 5.5 pounds at birth

(b) You choose three babies at random. What is the probability that their average birth weight is less than 5.5 pounds?

[tex]n = 3, s = \frac{1.25}{\sqrt{3}} = 0.7217[/tex]

This is the pvalue of Z when X = 5.5. So

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

By the Central Limit Theorem

This is the pvalue of Z when X = 5.5. So

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

[tex]Z = \frac{5.5 - 7.5}{0.7217}[/tex]

[tex]Z = -2.77[/tex]

[tex]Z = -2.77[/tex] has a pvalue of 0.0028

0.28% probability that their average birth weight is less than 5.5 pounds

Answer:

a) P ( 3 ≤X≤ 5 ) = 0.02619

b) E(X) = 1

Step-by-step explanation:

Given:

- The CDF of a random variable X = { 0 , 1 , 2 , 3 , .... } is given as:

                    [tex]F(X) = P ( X =< x) = 1 - \frac{1}{(x+1)*(x+2)}[/tex]

Find:

a.Calculate the probability that 3 ≤X≤ 5

b) Find the expected value of X, E(X), using the fact that. (Hint: You will have to evaluate an infinite sum, but that will be easy to do if you notice that

Solution:

- The CDF gives the probability of (X < x) for any value of x. So to compute the P (  3 ≤X≤ 5 ) we will set the limits.

                   [tex]F(X) = P ( 3=<X =< 5) = [1 - \frac{1}{(x+1)*(x+2)}]\limits^5_3\\\\F(X) = P ( 3=<X =< 5) = [-\frac{1}{(5+1)*(5+2)} + \frac{1}{(3+1)*(3+2)}}\\\\F(X) = P ( 3=<X =< 5) = [-\frac{1}{(42)} + \frac{1}{(20)}}]\\\\F(X) = P ( 3=<X =< 5) = 0.02619[/tex]

- The Expected Value can be determined by sum to infinity of CDF:

                   E(X) = Σ ( 1 - F(X) )

                   [tex]E(X) = \frac{1}{(x+1)*(x+2)} = \frac{1}{(x+1)} - \frac{1}{(x+2)} \\\\= \frac{1}{(1)} - \frac{1}{(2)}\\\\= \frac{1}{(2)} - \frac{1}{(3)} \\\\= \frac{1}{(3)} - \frac{1}{(4)}\\\\= ............................................\\\\= \frac{1}{(n)} - \frac{1}{(n+1)}\\\\= \frac{1}{(n+1)} - \frac{1}{(n+ 2)}[/tex]

                   E(X) = Limit n->∞ [1 - 1 / ( n + 2 ) ]  

                   E(X) = 1

Answer:

a. H + K = {u + v}

b. See explanation below

Step-by-step explanation:

Given

H+K={w:w=u+v for some u in H and some v in K}

a.

From the (given) above,

We have that

0 = v1 + 0.u

v1 represents the zero vector in K

We also have that

0 = v + u1

u1 represents the zero vector in H

Up this point, we've shown that both H and K have vector factors, V

We'll use the available data to solve for the zero vector of H + K

This is given by

u1 + v1 + u2 + v2 ---- Rearrange

u1 + u2 + v1 + v2 ---- Let u3 = u1 + u2 and v3 = v1 + v2.

So, we have

u3 + v3 as the zero vector of H + K

H + K = {u + v}

So, H + K is a subspace of V

b

Since H, K are subspace of V, then they are

1. Both close under scalar multiplication

2. Both closed under vector addition

1 and 2 gives

c(u + v) where c is a constant

= c.u + c.v

We can then conclude that H is a subspace of H+K and K is a subspace of H+K because H + K is closed under scalar multiplication and vector addition

The sum of two subspaces H and K is a subspace of V as it contains the zero vector, is closed under addition, and under scalar multiplication. Additionally, H and K themselves also qualify as subspaces of this resultant sum subspace, H+K.

To show that H+K is a subspace of V, we need to prove that it satisfies three properties: it contains the zero vector, it is closed under addition, and it is closed under scalar multiplication.

1. Zero Vector: Since H and K are subspaces, they both contain the zero vector. So, 0 = 0 + 0 represents a vector in H+K (with the first 0 from H and the second from K).

2. Closure under Addition: If w1 and w2 are vectors in H+K, w1 can be written as h1 + k1 (h1 in H, k1 in K) and w2 as h2 + k2 (h2 in H, k2 in K). Adding w1 and w2 gives: (h1 + k1) + (h2 + k2) = (h1 + h2) + (k1 + k2), where h1+h2 is in H and k1+k2 is in K, hence, the result is in H+K.

3. Closure under Scalar Multiplication: Let w be in H+K and c be any scalar. Then w = h + k, so cw = c(h + k) = ch + ck, where ch is in H and ck is in K, hence, cw is in H+K.

To show that H is a subspace of H+K and K is a subspace of H+K, note that for any vector h in H, it can be written as h+0 (with 0 as the zero vector of K), and similarly for any vector k in K as 0+k. Thus, H and K are subspaces of H+K.

https://brainly.com/question/30031077

#SPJ11

Answer:

The 99% confidence interval for the population mean hours spent watching television per month is between 143.07 hours and 158.93 hours.

Step-by-step explanation:

We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:

[tex]\alpha = \frac{1-0.99}{2} = 0.005[/tex]

Now, we have to find z in the Ztable as such z has a pvalue of [tex]1-\alpha[/tex].

So it is z with a pvalue of [tex]1-0.005 = 0.995[/tex], so [tex]z = 2.575[/tex]

Now, find M as such

[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]

In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.

[tex]M = 2.575*\frac{32}{\sqrt{108}} = 7.93[/tex]

The lower end of the interval is the mean subtracted by M. So it is 151 - 7.93 = 143.07 hours

The upper end of the interval is the mean added to M. So it is 151 + 7.93 = 158.93 hours

The 99% confidence interval for the population mean hours spent watching television per month is between 143.07 hours and 158.93 hours.

the answer is in the attachment

Answer:

Below the classical model, economic growth is necessarily achieved because of stability in the wage level. For instance, one case of unemployment predominates at a real wage (W / P)1.

Currently, the excessive labor supply would lower the actual wage level before labor supply equals its demand. Eventually, real wage rates would decline to (W / P)F, whereby aggregate labor demand is perfectly matches by aggregate labor supply.

Only the supply side of the production market for products defines the quantity of output & jobs in the classical model.

As the classical method is supply-determined, it states that equiproportional increases (or declines) will not alter the supply of labor in both the rate of money wage and the price level.

Answer: the probability that a measurement exceeds 13 milliamperes is 0.067

Step-by-step explanation:

Suppose that the current measurements in a strip of wire are assumed to follow a normal distribution, we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = current measurements in a strip.

µ = mean current

σ = standard deviation

From the information given,

µ = 10

σ = 2

We want to find the probability that a measurement exceeds 13 milliamperes. It is expressed as

P(x > 13) = 1 - P(x ≤ 13)

For x = 13,

z = (13 - 10)/2 = 1.5

Looking at the normal distribution table, the probability corresponding to the z score is 0.933

P(x > 13) = 1 - 0.933 = 0.067

Final answer:

To find the probability that a current measurement exceeds 13 milliamperes in a normally distributed set with mean 10 mA and SD 2 mA, calculate the Z-score and use a normal distribution table or software.

Explanation:

The student's question seems to mistakenly mix units (millimeters instead of milliamperes), but assuming the intent was to refer to electrical current and not physical measurements of wire, we will proceed on the basis that the actual question is about the probability of a current measurement exceeding 13 milliamperes.

To calculate the probability that a measurement exceeds 13 milliamperes when the current measurements in a strip of wire are normally distributed with a mean of 10 milliamperes and standard deviation of 2 milliamperes, we need to use the Z-score formula:

Z = (X - μ) / σ

where X is the value in question (13 milliamperes), μ is the mean (10 milliamperes), and σ is the standard deviation (2 milliamperes).

Plugging in the values:

Z = (13 - 10) / 2 = 1.5

After finding the Z-score, we would look up this value in a standard normal distribution table or use a statistical software to find the probability that Z exceeds 1.5, which gives us the probability that a measurement exceeds 13 milliamperes. For this Z-score, the probability is approximately 6.68%.

Answer: 60 gallons of the 20% solution and 120 gallons of the 50% solution should be mixed.

Step-by-step explanation:

Let x represent the number of gallons of 20% solution that should be mixed.

Let y represent the number of gallons of 50% solution that should be mixed.

A 20% solution of fertilizer is to be mixed with a 50% solution of fertilizer in order to get 180 gallons of a 40% solution. This means that

0.2x + 0.5y = 0.4×180

0.2x + 0.5y = 72- - - - - - - - - - - -1

Since the total number of gallons is 180, it means that

x + y = 180

Substituting x = 180 - y into equation 1, it becomes

0.2(180 - y) + 0.5y = 72

36 - 0.2y + 0.5y = 72

- 0.2y + 0.5y = 72 - 36

0.3y = 36

y = 36/0.3

y = 120

x = 180 - y = 180 - 120

x = 60

To get a 40% solution by mixing a 20% and 50% solution, you need 60 gallons of the 20% solution and 120 gallons of the 50% solution.

To solve this problem, we can use the concept of mixtures. Let's represent the amount of 20% solution as x gallons and the amount of 50% solution as y gallons. From the given information, we can set up the following equations:

x + y = 180 (equation 1)

0.2x + 0.5y = 0.4 * 180 (equation 2)

Simplifying equation 2 gives us 0.2x + 0.5y = 72. To eliminate decimals, we can multiply both sides by 10 to get 2x + 5y = 720.

Now we have a system of equations. We can solve it by substitution or elimination. Let's use elimination:

Multiplying equation 1 by 2 gives us 2x + 2y = 360. Subtracting this from equation 2 gives us 3y = 360. Dividing both sides by 3 gives us y = 120. Substituting this value into equation 1 gives us x + 120 = 180. Subtracting 120 from both sides gives us x = 60.

Therefore, we need 60 gallons of the 20% solution and 120 gallons of the 50% solution.

https://brainly.com/question/22742069

#SPJ3

Answer:

[tex]\dfrac{d}{dx}f(x)=\dfrac{e^{2x}+e^{-2x}}{2}[/tex]

Step-by-step explanation:

It is given that

[tex]\sinh x=\dfrac{e^x-e^{-x}}{2}[/tex]

[tex]\cosh x=\dfrac{e^x+e^{-x}}{2}[/tex]

[tex]f(x)=\sinh x\cosh x=[/tex]

Using the given hyperbolic functions, we get

[tex]f(x)=\dfrac{e^x-e^{-x}}{2}\times \dfrac{e^x+e^{-x}}{2}[/tex]

[tex]f(x)=\dfrac{(e^x)^2-(e^{-x})^2}{4}[/tex]

[tex]f(x)=\dfrac{e^{2x}-e^{-2x}}{4}[/tex]

Differentiate both sides with respect to x.

[tex]\dfrac{d}{dx}f(x)=\dfrac{d}{dx}\left(\dfrac{e^{2x}-e^{-2x}}{4}\right )[/tex]

[tex]\dfrac{d}{dx}f(x)=\left(\dfrac{2e^{2x}-(-2)e^{-2x}}{4}\right )[/tex]

[tex]\dfrac{d}{dx}f(x)=\left(\dfrac{2(e^{2x}+e^{-2x})}{4}\right )[/tex]

[tex]\dfrac{d}{dx}f(x)=\dfrac{e^{2x}+e^{-2x}}{2}[/tex]

Hence, [tex]\dfrac{d}{dx}f(x)=\dfrac{e^{2x}+e^{-2x}}{2}[/tex].

Answer:

Complete question for your reference is below.

A researcher interested in weight control wondered whether normal and overweight individuals differ in their reaction to the availability of food. Thus, normal and overweight participants were told to eat as many peanuts as they desired while working on a questionnaire. One manipulation was the proximity of the peanut dish (close or far from the participant); the second manipulation was whether the peanuts were shelled or unshelled. After filling out the questionnaire, the peanut dish was weighed to determine the amount of peanuts consumed.

1. Identify the design (e.g., 2 X 2 factorial).

2. Identify the total number of conditions.

3. Identify the manipulated variable(s).

4. Is this an IV X PV design? If so, identify the participant variable(s).

5. Is this a repeated measures design? If so, identify the repeated variable(s).

6. Identify the dependent variable(s).

Step-by-step explanation:

Please find attached file for complete answer solution and explanation.

The research design is a 2x2 factorial with the manipulated or independent variables being the proximity of the peanut dish and whether the peanuts are shelled, resulting in four conditions. It is not an IV x PV design, nor a repeated measures design. The dependent variable in this study is the amount of peanuts consumed.

The research design posed in this question appears to be a 2x2 factorial design. This is because there are two independent variables (the proximity of the peanut dish and whether the peanuts are shelled or not), each with two levels (close or far, and shelled or unshelled respectively).

Given this, there would be a total of four conditions in this study (close-shelled, close-unshelled, far-shelled, far-unshelled). The manipulated variables, also referred to as independent variables, in this study are the 'proximity of the peanut dish' and 'whether the peanuts are shelled or unshelled'.

This does not appear to be an IV X PV (Independent Variable x Participant Variable) design because there is no information about a participant variable being used here. It also does not seem to be a repeated measures design as individuals are not exposed to all conditions; rather they experience one specific condition. The dependent variable is the amount of peanuts consumed, as determined by weighing the peanut dish after participants have finished eating.

https://brainly.com/question/28039772

#SPJ11

Answer:

0.16

Step-by-step explanation:

For a normal distribution, the 68-95-99.7 rule says that 68% of the distribution lies within 1 standard deviation from the mean, 95% within two standard deviations and 99.7% within three standard deviations.

From the question,

Mean μ = 298 ml

Standard deviation σ = 3 ml

A value of 295 ml is within one standard deviation = 68%.

Since it's on the lower side, it's within 68% ÷ 2 = 34%.

The lower half below the mean is 50% of the distribution. Hence, for a selection less than 1 standard deviation, the probability is

50% - 34% = 16% = 0.16

Final answer:

The probability that a randomly selected bottle contains less than 295 ml of cola is calculated using the normal distribution. With a z-score of -1, ths probability is approximately 15.87%.

Explanation:

To determine the probability that a randomly selected bottle contains less than 295 ml of cola, we can use the properties of the normal distribution.

Given that the mean is 298 ml and the standard deviation is 3 ml, we must first find the z-score associated with 295 ml.

The z-score formula is z = (X - mu) / SD where X is the value for which we want to find the probability.

Substituting the given values into the formula, we get:

z = (295 - 298) / 3 = -1.

A z-score of -1 tells us that 295 ml is one standard deviation below the mean.

To find the probability corresponding to this z-score, we refer to a standard normal distribution table or use relevant statistical software. If we look up the probability for z = -1 in the z-table, we find that the area to the left of z is approximately 0.1587.

Therefore, the probability that a randomly selected bottle contains less than 295 ml of cola is about 15.87%.

The Main Answer for:

1. The value of the variable for which the trinomial [tex]a^2+7a+6[/tex] and the binomial a+1 have the same value is [tex]a=-1[/tex] or [tex]a=-5[/tex].

2. The value of the variable for which the trinomial [tex]3x^2-x+1[/tex] and the trinomial [tex]2x^2+5x-4[/tex] have the same value is [tex]x=1[/tex] or [tex]x=5[/tex].

To find the value of the variable for which two polynomials have the same value, we set them equal to each other and solve for the variable.

1. For the trinomial [tex]a^2+7a+6[/tex] and the binomial [tex]a+1[/tex]:

[tex]a^2+7a+6=a+1[/tex]

First, let's rewrite the equation in standard form:

[tex]a^2+7a+6-(a+1)=0\\a^2+7a+6-a-1=0\\a^2+6a+5=0[/tex]

Now, we can solve this quadratic equation by factoring:

[tex]a^2+6a+5=0\\(a+5)(a+1)=0[/tex]

Setting each factor equal to zero:

[tex]a+5=0[/tex]⇒[tex]a=-5[/tex]

[tex]a+1=0[/tex]⇒[tex]a=-1[/tex]

So, the value of the variable for which the trinomial [tex]a^2+7a+6[/tex] and the binomial a+1 have the same value is [tex]a=-1[/tex] or [tex]a=-5[/tex].

2. For the trinomial [tex]3x^2-x+1[/tex] and the trinomial [tex]2x^2+5x-4[/tex]:

[tex]3x^2-x+1=2x^2+5x-4[/tex]

Again, let's rewrite the equation in standard form:

[tex]3x^2-x+1-(2x^2+5x-4)=0\\3x^2-x+1-2x^2-5x+4=0\\X^2-6x+5=0[/tex]

Now, let's solve this quadratic equation by factoring:

[tex]X^2-6x+5=0\\(x-5)(x-1)=0[/tex]

Setting each factor equal to zero:

[tex]x-5=0[/tex]⇒[tex]x=5[/tex]

[tex]x-1=0[/tex]⇒[tex]x=1[/tex]

So, the value of the variable for which the trinomial [tex]3x^2-x+1[/tex] and the trinomial [tex]2x^2+5x-4[/tex] have the same value is [tex]x=1[/tex] or [tex]x=5[/tex].

COMPLETE QUESTION:

Find out for what value of the variable:

1. Do the trinomial [tex]a^2+7a+6[/tex] and the binomial [tex]a+1[/tex] have the same value?

2. Do the trinomial [tex]3x^2-x+1[/tex] and the trinomial [tex]2x^2+5x-4[/tex] have the same value?