A soccer ball manufacturer wants to estimate the mean circumference of soccer balls within 0.1 in. Determine the minimum sample size required to construct a 95% confidence interval for the population mean. Assume the population standard deviation is 0.25 in.

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:

There is enough evidence to support the claim that the average number of sheets recycled per bin was more than 59.3 sheets.          

Step-by-step explanation:

We are given the following in the question:  

Population mean, μ = 59.3

Sample mean, [tex]\bar{x}[/tex] = 62.4

Sample size, n = 79

Alpha, α = 0.05

Sample standard deviation, s = 9.86

First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu = 59.3\text{ sheets}\\H_A: \mu > 59.3\text{ sheets}[/tex]

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

Formula:

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

Putting all the values, we have

[tex]t_{stat} = \displaystyle\frac{62.4 - 59.3}{\frac{9.86}{\sqrt{79}} } = 2.7945[/tex]

Degree of freedom = n - 1 = 78

Now, [tex]t_{critical} \text{ at 0.05 level of significance, 78 degree of freedom } = 1.6646[/tex]

Since,                        

[tex]t_{stat} > t_{critical}[/tex]

We fail to accept the null hypothesis and reject it. We accept the alternate hypothesis.

Conclusion:

Thus, there is enough evidence to support the claim that the average number of sheets recycled per bin was more than 59.3 sheets.

Answer:

[tex]r=\frac{19M}{14h}[/tex]

Step-by-step explanation:

The equation is given as:

[tex]M=\frac{7r2h}{19}[/tex]

Assuming all the unknown variables are positive, we can make [tex]r[/tex] the subject of the formula to obtain it in terms of M & h:

[tex]M=\frac{7r2h}{19}\\M\times19=7r2h\\\\\frac{19M}{2h}=7r\\\\r=\frac{19M}{2h\times7}\\\\r=\frac{19M}{14h}[/tex]

or [tex]r=1.3571M/h[/tex]

Hence, r as in terms of M& H is given as

[tex]r=\frac{19M}{14h} \ or \ 1.3571M/h[/tex]

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.

Answer:

You earned an A.

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 = 79, \sigma = 6[/tex]

The top 15% of all scores have been designated As.

This means that if Z for the score has a pvalue of 1-0.15 = 0.85 or higher, the score is designated as A.

Your score is 89. Did you earn an A?

We have to find the pvalue of Z when X = 89. So

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

[tex]Z = \frac{89 - 79}{6}[/tex]

[tex]Z = 1.67[/tex]

[tex]Z = 1.67[/tex] has a pvalue of 0.9525. So yes, you earned an A.

Answer:

The answer is a population parameter.

Step-by-step explanation:

Population can include people, but other examples include objects, event, businesses, and so on. Population is the entire pool from which statistical sample is drawn.

A parameter is a value that describes a characteristics of an entire population, such as population mean, because you can almost never measure an entire population, you usually don't know the real value of a parameter.

Consider all possible sample of size N that can be drawn from a given population (either with or without replacement). For example, we can compute a statistics (such as the mean and the standard deviation ) that will vary from sample to sample. In this manner we obtain a distribution of statistics that is called Sampling distribution.

In statistics, a sampling distribution is the theoretical distribution of a sample statistic that arises from drawing all possible samples of a specific size from a population. It helps to quantify the variability and predictability of sample statistics when used as estimates for population parameters.

A sampling distribution refers to the "distribution of a sample statistic". This is option C from your list. This term describes the probability distribution of a statistic based on a random sample. For example, if we study random samples of a certain size from any population, the mean score will form a distribution. This is the sampling distribution of the mean. Similarly, variance, standard deviations and other statistics also have sampling distributions. The purpose of a sampling distribution is to quantify the variation and uncertainty that arises when we use sample statistics (like the mean) to estimate population parameters (like the population mean).

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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:

hope it helps you see the attachment for further information

Answer:

81

Step-by-step explanation:

Answer:

a) Mean = 1.6, standard deviation = 1.21

b) 18.87% probability that zero online retail orders will turn out to be fraudulent.

c) 32.82% probability that one online retail order will turn out to be fraudulent.

d) 48.31% probability that two or more online retail orders will turn out to be fraudulent.

Step-by-step explanation:

Binomial probability distribution

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

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

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

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

And p is the probability of X happening.

The mean 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]

In this problem, we have that:

[tex]p = 0.08, n = 20[/tex]

a. What are the mean and standard deviation of the number of online retail orders that turn out to be fraudulent?

Mean

[tex]E(X) = np = 20*0.08 = 1.6[/tex]

Standard deviation

[tex]\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{20*0.08*0.92} = 1.21[/tex]

b. What is the probability that zero online retail orders will turn out to be fraudulent?

This is P(X = 0).

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

[tex]P(X = 0) = C_{20,0}.(0.08)^{0}.(0.92)^{20} = 0.1887[/tex]

18.87% probability that zero online retail orders will turn out to be fraudulent.

c. What is the probability that one online retail order will turn out to be fraudulent?

This is P(X = 1).

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

[tex]P(X = 1) = C_{20,1}.(0.08)^{1}.(0.92)^{19} = 0.3282[/tex]

32.82% probability that one online retail order will turn out to be fraudulent.

d. What is the probability that two or more online retail orders will turn out to be fraudulent?

Either one or less is fraudulent, or two or more are. The sum of the probabilities of these events is decimal 1. So

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

We want [tex]P(X \geq 2)[/tex]

So

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

In which

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

From itens b and c

[tex]P(X \leq 1) = 0.1887 + 0.3282 = 0.5169[/tex]

[tex]P(X \geq 2) = 1 - P(X \leq 1) = 1 - 0.5169 = 0.4831[/tex]

48.31% probability that two or more online retail orders will turn out to be fraudulent.

The probability is an illustration of a binomial distribution.

The given parameters are:

n = 20

p = 0.08

The mean is calculated as:

[tex]\bar x = np[/tex]

So, we have:

[tex]\bar x = 20 * 0.08[/tex]

[tex]\bar x = 1.6[/tex]

The standard deviation is calculated as:

[tex]\sigma = \sqrt{\bar x * (1 - p)[/tex]

This gives

[tex]\sigma = \sqrt{1.6 * (1 - 0.08)[/tex]

[tex]\sigma = 1.21[/tex]

Hence, the mean is 1.6 and the standard deviation is 1.21

This is calculated as:

[tex]P(x) = ^nC_x * p^x * (1 - p)^{n-x}[/tex]

So, we have:

[tex]P(0) = ^{20}C_0 * 0.08^0 * (1 - 0.08)^{20 - 0}[/tex]

[tex]P(0) =0.1887[/tex]

The probability that zero online retail orders will turn out to be fraudulent is 0.1887

This is calculated as:

[tex]P(x) = ^nC_x * p^x * (1 - p)^{n-x}[/tex]

So, we have:

[tex]P(1) = ^{20}C_1 * 0.08^1 * (1 - 0.08)^{20 - 1}[/tex]

[tex]P(1) =0.3281[/tex]

The probability that one online retail orders will turn out to be fraudulent is 0.3281

This is calculated as:

[tex]P(x\ge 2) = 1 - P(0) - P(1)[/tex]

So, we have:

[tex]P(x\ge 2) = 1 - 0.1887 - 0.3281[/tex]

[tex]P(x\ge 2) = 0.4832[/tex]

The probability that two or more online retail orders will turn out to be fraudulent is 0.4832

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Answer:

The 95% confidence interval for the average heights of the population is between 1.7108m and 1.7892m.

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{1-0.95}{2} = 0.025[/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]1-0.025 = 0.975[/tex], so [tex]z = 1.96[/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.

The standard deviation is the square root of the variance. So

[tex]\sigma = \sqrt{0.16} = 0.4[/tex]

Then

[tex]M = 1.96*\frac{0.4}{\sqrt{400}} = 0.0392[/tex]

The lower end of the interval is the mean subtracted by M. So it is 1.75 - 0.0392 = 1.7108m

The upper end of the interval is the mean added to M. So it is 1.75 + 0.0392 = 1.7892m

The 95% confidence interval for the average heights of the population is between 1.7108m and 1.7892m.

Answer:

Step-by-step explanation:

mean x = [tex]1.75[/tex]

Variance [tex]\rho^2 = 0.16[/tex]

standard deviation [tex](\rho) = \sqrt{0.16} = 0.4[/tex]

n = 400

[tex]95\%[/tex] confidence :

[tex]\alpha = 100\% - 95\% = 5\%\\\\\frac{\alpha}{2} = 2.5\% = 0.025[/tex]

From standard normal distribution table,

[tex]Z_\frac{\alpha}{2} = Z_{0.025} = 1.96[/tex]

Margin of error, [tex]E = Z_\frac{\alpha}{2} * \frac{\rho}{\sqrt{n}}[/tex]

[tex]E = 1.96 * \frac{0.4}{\sqrt{400}}\\\\E = 0.0392[/tex]

Lower limit: x - E

[tex]= 1.75 - 0.0392\\\\= 1.7108[/tex]

Upper limit: x + E

[tex]= 1.75 + 0.0392\\\\= 1.7892[/tex]

[tex]Limits : (1.7108, 1.7892)[/tex]

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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.

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Answer:

True for n = 18, 19, 20, 21

Step-by-step explanation:

[tex]P(n) =[/tex] a postage of [tex]n[/tex] cents; where [tex]P(n) = 4x + 7y[/tex]. ( [tex]x[/tex] are the number of 4-cent stamps and [tex]y[/tex] are the number of 7-cent stamps)

For [tex]n=18, P(18)[/tex] is true.

This is a possibility, if [tex]x= 1 \ and \ y=2[/tex]

[tex]P(18) = 4(1) + 7(2) = 4 + 14 = 18[/tex]

Similarly for [tex]P(19)[/tex]:

[tex]P(19) = 4(3) + 7(0) = 19[/tex]

[tex]P(20) = 4(5) + 7(0) = 20\\P(21) = 4(0) + 7(3) = 21[/tex]

Answer:

Step-by-step explanation:

A quadrilateral inscribed within a circle is known as a cyclic quadrilateral. Property of the cyclic quadrilateral is that, sum of its opposite angles is 180°.

∴ ∠U + ∠K = 180°

∴ ∠K = 180 - 85 = 95°

Answer and Step-by-step explanation:

The answer is attached below

In this exercise we have to use the knowledge of parameterization and calculate the direction and direction of the equation, so we have to:

A) Clockwise: [tex]t \in [ -\infty, 1/6][/tex]

B) Counter-clockwise: [tex]t \in [ 1/6, \infty][/tex]

C) [tex]\theta \in [ 0, 2 \pi][/tex]

D) [tex]t= 0 \ or \ t=1/3[/tex]

For this exercise, the following equations were informed:

[tex]x= cos(3t^2-t)\\y= sin(3t^2-t)\\t \in [ -\infty, \infty][/tex]

taking the parameterization we have that:

[tex]\phi = 3t^2 - t= t(3t-1)[/tex]

As t increases from [tex][ -\infty, \infty][/tex]  [tex]\phi[/tex] decreases, after 0 it becomes negative and after 1/3, goes on increasing. Also:

[tex]\frac{d\phi}{dt} = (6t-1)\\t= 1/6[/tex]

a) For clockwise begin [tex]\phi[/tex] must be decreasing, so:

[tex]t \in [ -\infty, 1/6][/tex]

b) For counter-clockwise  [tex]\phi[/tex] must be increasing, so:

[tex]t \in [ 1/6, \infty][/tex]

c) Entise circle gets traced out. For we know:

[tex]x= cos\theta\\y= sin\theta[/tex]

Circle gets traced out once for:

[tex]\theta \in [ 0, 2 \pi][/tex]

d) When point (1, 0) so:

[tex]1= cos(3t^2-t)\\0= sin(3t^2-t)\\t= 0 \or \ t=1/3[/tex]

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Answer:

[tex] P(X=1)[/tex]

And using the probability mass function we got:

[tex] P(X=1) = (5C1) (0.25)^1 (1-0.25)^{5-1}= 0.396[/tex]

Step-by-step explanation:

Previous concepts  

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".  

Solution to the problem  

For this cae that one buggy whip would be defective is [tex] p = \frac{5}{20}=0.25[/tex]

Let X the random variable of interest, on this case we now that:  

[tex]X \sim Binom(n=5, p=0.25)[/tex]  

The probability mass function for the Binomial distribution is given as:  

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

Where (nCx) means combinatory and it's given by this formula:  

[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]  

And we want to find this probability:

[tex] P(X=1)[/tex]

And using the probability mass function we got:

[tex] P(X=1) = (5C1) (0.25)^1 (1-0.25)^{5-1}= 0.396[/tex]

The probability that a randomly selected Canadian baby is a large baby (weighing more than 4,000 grams) is approximately 0.187 or 18.7%.

To find the probability that a randomly selected Canadian baby is a large baby, we need to calculate the area under the normal distribution curve to the right of 4,000 grams. First, we calculate the z-score using the formula: z = (x - mean) / standard deviation. Plugging in the values, we get z = (4000 - 3500) / 560 = 0.8929.

Next, we need to find the area under the curve to the right of this z-score using a standard normal distribution table or a calculator. The cumulative probability from the table or calculator is approximately 0.187. This means that the probability of a randomly selected Canadian baby being a large baby (weighing more than 4,000 grams) is approximately 0.187 or 18.7%.

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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:

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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)f(t)=1090-5t

(b)f(0)=1090metres

(c)g(t)=1600-9t

(d)g(0)=1600metres

(e)The Hare

Step-by-step explanation:

Total Distance =1600 metres

(a)The tortoise has a 510m headstart and a speed of 5m/s

Distance=Speed X Time

Distance at 5m/s = 5t

Total Distance covered by the tortoise at any time t= 510+5t

Therefore, The tortoise's distance from the finish line (in meters) in terms of the number of seconds t since the start of the race is given as:

f(t)=1600-(510+5t)

f(t)=1090-5t

(b)f(0)=1090-(5X0)

=1090metres

(c)The hare has a speed of 9m/s

Distance=Speed X Time

Distance at 9m/s = 9t

Total Distance covered by the hare at any time t= 9t

Therefore, The hare's distance from the finish line (in meters) in terms of the number of seconds t since the start of the race is given as:

g(t)=1600-9t

(d)

g(0)=1600-(9X0)=1600metres

(e)The race is finished when the distance from the finish line=0

For the Tortoise

f(t)=1090-5t=0

1090=5t

t=218seconds

For the Hare

g(t)=1600-9t=0

9t=1600

t=177.8seconds

The hare takes a shorter time to reach the finish line so he won the race.

Answer:

The function is provided below.

Step-by-step explanation:

Running the program in Python, the code for this problem is as follows:

def func(n):

       count = 0                                           **initializing the sum with 0**

       for i in range (1, n):

                   if i%3 == 0 or i%5 == 0:        

                              count = count + 1

      return (count)

When the program is executed enter value 1 to 10 one by one and the result will be 23, the sum of all the multiples of 3 and 5.

Answer: 31

Step-by-step explanation:

This is solved Using the theory of sets.

Teachers in Albers school = 44

Teachers in Bothel school = 74

Let the number of teachers that work in both schools be denoted as "x"

This implies that:

Number of teachers in Albers only = 44 - x

Number of teachers in Bothel only = 74 - x

And total number of teachers in both schools = 87,

then

x + (44-x) + (74-x) = 87

118 -x = 87

31 = x

This means the number of teachers that teach in both schools = 31.

Answer: 31 teachers teach at both schools.

Let:

A =  Albers Elementary School

B = Bothel Elementary School

According to the question:

n(A) = 44, n(B) = 74, n(A U B) = 87.

Using the union formula we get:

[tex]n(A \cup B) = n(A)+n(B)-n(A \cap B)\\87 = 44+74-n(A \cap B)\\n(A \cap B)=44+74-87\\n(A \cap B)=31[/tex]

So, 31 teachers teach at both schools.

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Answer:

[tex]f(t)=1804(1.04)^{t}\\f(25)=4809.17[/tex]

Step-by-step explanation:

1. Since the increasing rate is 0.04 or (4%) per day, then the factor is (1+0.04) raised to t days, and we have and exponential growth therefore we can write:

[tex]\\f(t)=c(1.04)^t\\f(t)=1,804(1.04)^t\\[/tex]

2. To estimate the number of cases, 25 days later following that exponential model

[tex]f(25)=1804(1.04)^{25}\\f(25)=4809.17[/tex]

Answer:

11.51% probability of observing an 83 degree F temperature or higher LA during a randomly chosen day in June

The coldest 10% of the days during June in LA have high temperatures of 70.6F or lower.

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 = 77, \sigma = 5[/tex]

What is the probability of observing an 83 degree F temperature or higher LA during a randomly chosen day in June?

This probability is 1 subtracted by the pvalue of Z when X = 83. So

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

[tex]Z = \frac{83 - 77}{5}[/tex]

[tex]Z = 1.2[/tex]

[tex]Z = 1.2[/tex] has a pvalue of 0.8849.

1 - 0.8849 = 0.1151

11.51% probability of observing an 83 degree F temperature or higher LA during a randomly chosen day in June

How cold are the coldest 10% of the days during June in LA?

High temperatures of X or lower, in which X is found when Z has a pvalue of 0.1, so whn Z = -1.28

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

[tex]-1.28 = \frac{X - 77}{5}[/tex]

[tex]X - 77 = -1.28*5[/tex]

[tex]X = 70.6[/tex]

The coldest 10% of the days during June in LA have high temperatures of 70.6F or lower.

A) The probability of observing a temperature ≥ 83°F in LA during a randomly chosen day in June is;

p(observing a temperature ≥ 83°F) = 11.507%

B) The coldest 10% of the days during June in LA have temperatures;

Less than or equal to 70.592 °F

This question involves z-distribution which is given by the formula;

z = (x' - μ)/σ

We are given;

Average daily temperature; μ = 77 °F

Standard deviation; σ = 5 °F

Since the temperatures follow a normal distribution, then if we want to find the probability of observing a temperature ≥ 83°F, then;

x' = 83 °F

Thus;

z = (83 - 77)/5

z = 6/5

z = 1.2

Thus;

from online z-score calculator, p-value = 0.11507

Thus, p(observing a temperature ≥ 83°F) = 11.507%

B) We want to find out how cold the coldest 10% of the days during June in LA;

Thus, it means that p = 10% = 0.1

z-score at p = 0.1 from z-score tables is;

z = -1.28155

Thus;

-1.28155 = (x' - 77)/5

-1.28155*5 = x' - 77

-6.40775 = x' - 77

x' = 77 - 6.40775

x' ≈ 70.592 °F

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Follow below steps:

To solve for q in the equation √3q + 2 = √5, we first isolate the term with q by subtracting 2 from both sides of the equation.

√3q = √5 - 2

Then we square both sides of the equation to remove the square root:

(√3q)² = ( √5 - 2 )²

3q = ( √5 - 2 )²

Now, expand the right side of the equation:

3q = 5 - 2√5 * 2 + 2²

3q = 5 - 4√5 + 4

3q = 9 - 4√5

Then, divide both sides of the equation by 3 to solve for q:

q = (9 - 4√5) / 3

So, the value of q is (9 - 4√5) / 3.

The value of q is [tex]\frac{9 - 4\sqrt{5}}{3}[/tex].

To solve for q, we'll isolate it by performing operations to both sides of the equation to get q by itself.

Given the equation:

[tex]\[ \sqrt{3q} + 2 = \sqrt{5} \][/tex]

Subtract 2 from both sides:

[tex]\[ \sqrt{3q} = \sqrt{5} - 2 \][/tex]

Now, to isolate q, we need to square both sides of the equation:

[tex]\[ (\sqrt{3q})^2 = (\sqrt{5} - 2)^2 \]\[ 3q = (\sqrt{5} - 2)^2 \]\[ 3q = 5 - 4\sqrt{5} + 4 \]\[ 3q = 9 - 4\sqrt{5} \][/tex]

Now, divide both sides by 3 to solve for q:

[tex]\[ q = \frac{9 - 4\sqrt{5}}{3} \][/tex]

So, [tex]\( q = \frac{9 - 4\sqrt{5}}{3} \).[/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.

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Answer:

There are 21 comparisons to be made.

Step-by-step explanation:

The number of species of Odonata monitored every year is, n = 7.

It is provided that the density of each species is compared with each other.

The number of ways to compare the species (N) without repetition is:

[tex]N=\frac{n(n-1)}{2}\\=\frac{7(7-1)}{2}\\=\frac{7\times6}{2}\\=21[/tex]

Thus, there are 21 comparisons.

The 21 comparisons are as follows:

Specie 1 is compared with the remaining 6.

Specie 2 has already with he 1st so it is compared with the remaining 5.

Specie 3 has already with he 1st and 2nd so it is compared with the remaining 4.

Specie 4 has already with he 1st, 2nd and 3rd so it is compared with the remaining 3.

Specie 5 has already with he 1st, 2nd, 3rd and 4th so it is compared with the remaining 2.

Specie 6 has already with he 1st, 2nd, 3rd, 4th and 5th so it is compared with the remaining 1.

And the specie 7 has already been compared with the others.

Total number of comparisons = 6 + 5 + 4 + 3 + 2 + 1 = 21.

Answer:

[tex]ydv = (v +2)dy\\[/tex]            

Step-by-step explanation:

We are given the following differential equation:

[tex]y dx = 2(x + y) dy[/tex]

We have to substitute

[tex]x = vy[/tex]

Differentiating we get,

[tex]\dfrac{dx}{dy} = v + y\dfrac{dv}{dy}[/tex]

Putting value in differential equation, we get,

[tex]y dx = 2(x + y) dy\\\\y\dfrac{dx}{dy}=2(x+y)\\\\y(v+y\dfrac{dv}{dy}) = 2(vy + y)\\\\vy + y^2\dfrac{dv}{dy} = 2vy +2y\\\\y^2\dfrac{dv}{dy}=vy +2y\\\\y^2dv = y(v+2)dy\\ydv = (v +2)dy\\[/tex]

is the differential equation after substitution.

The given homogeneous differential equation y dx = 2(x + y) dy can be rewritten in terms of y and v using the substitution x = vy. The result is the differential equation y dv/dy = v.

The given differential equation is y dx = 2(x + y) dy. To write his equation in terms of y and v using the substitution x = vy, we must first differentiate both sides of x = vy with respect to x to get 1 = v + y dv/dx. We rearrange this to get dx/dy = 1 / (v + y dv/dy). The original equation can now be rewritten after substituting these values, you will get y / (v + y dv/dy) = 2(v + y), simplifying, we get v = 2v + 2y, and after rearranging, we get y dv/dy = v. This is the differential equation in terms of v and y.

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Answer:

4.76% probability of between 2 and 4 phone calls received (endpoints included) in a given day.

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 interval.

Records show that the average number of phone calls received per day is 9.2.

This means that [tex]\mu = 9.2[/tex].

Find the probability of between 2 and 4 phone calls received (endpoints included) in a given day.

[tex](2 \leq X \leq 4) = P(X = 2) + P(X = 3) + P(X = 4)[/tex]

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

[tex]P(X = 2) = \frac{e^{-9.2}*(9.2)^{2}}{(2)!} = 0.0043[/tex]

[tex]P(X = 3) = \frac{e^{-9.2}*(9.2)^{3}}{(3)!} = 0.0131[/tex]

[tex]P(X = 4) = \frac{e^{-9.2}*(9.2)^{4}}{(4)!} = 0.0302[/tex]

[tex](2 \leq X \leq 4) = P(X = 2) + P(X = 3) + P(X = 4) = 0.0043 + 0.0131 + 0.0302 = 0.0476[/tex]

4.76% probability of between 2 and 4 phone calls received (endpoints included) in a given day.

Answer:

Correct option: B. Less between group variability

Step-by-step explanation:

The Analysis of Variance (ANOVA) test is performed to determine whether there is a significant difference between the different group mean.

The hypothesis is defined as:

H₀: There is no difference between the group means, i.e. μ₁ = μ₂ = ... = μₙ

Hₐ: At least one of the mean is different from the others, i.e. μ[tex]_{i}[/tex] ≠ 0.

The test statistic is defined as:

[tex]F=\frac{SS_{between}}{SS_{within}}[/tex]

If the null hypothesis is true then the test statistic will be small and if it is false then the test statistic will be large.

In this case it is provided that the null hypothesis is true.

This implies that:

[tex]SS_{between}<SS_{within}[/tex]

Implying that the sum of squares for between group variability is less than within group variability.

Thus, if the null hypothesis is true there will be less between group variability.

Answer:

97.10% probability that five or more of the original 2000 components fail during the useful life of the product.

Step-by-step explanation:

For each component, there are only two possible outcomes. Either it works correctly, or it does not. The probability of a component falling is independent from other components. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

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

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

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

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

And p is the probability of X happening.

In this problem we have that:

[tex]n = 2000, p = 1-0.995 = 0.005[/tex]

Approximate the probability that five or more of the original 2000 components fail during the useful life of the product.

We know that either less than five compoenents fail, or at least five do. The sum of the probabilities of these events is decimal 1. So

[tex]P(X < 5) + P(X \geq 5) = 1[/tex]

We want [tex]P(X \geq 5)[/tex]

So

[tex]P(X \geq 5) = 1 - P(X < 5)[/tex]

In which

[tex]P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)[/tex]

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

[tex]P(X = 0) = C_{2000,0}.(0.005)^{0}.(0.995)^{2000} = 0.000044[/tex]

[tex]P(X = 1) = C_{2000,1}.(0.005)^{1}.(0.995)^{1999} = 0.000445[/tex]

[tex]P(X = 2) = C_{2000,2}.(0.005)^{2}.(0.995)^{1998} = 0.002235[/tex]

[tex]P(X = 3) = C_{2000,3}.(0.005)^{3}.(0.995)^{1997} = 0.007480[/tex]

[tex]P(X = 4) = C_{2000,4}.(0.005)^{4}.(0.995)^{1996} = 0.018765[/tex]

[tex]P(X < 5) = P(X = 0) + `P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.000044 + 0.000445 + 0.002235 + 0.007480 + 0.018765 = 0.0290[/tex]

[tex]P(X \geq 5) = 1 - P(X < 5) = 1 - 0.0290 = 0.9710[/tex]

97.10% probability that five or more of the original 2000 components fail during the useful life of the product.