Discrete or Continuous? Identify the random variables in Exercises 2, 3, 4, 5, 6, 7, 8, 9, 10, and 11 as either discrete or continuous. Total number of points scored in a football game

Answer:

[tex] t = 2.24[/tex]

The first step is calculate the degrees of freedom, on this case:  

[tex]df=n-1=23-1=22[/tex]  

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

[tex]p_v =P(t_{(22)}>2.24)=0.01776[/tex]  

And for this case we can conclude that:

[tex] 0.01 < p_v < 0.025[/tex]

And we will reject the null hypothesis at [tex] \alpha=0.025[/tex] since [tex] p_v < \alpha[/tex]

Step-by-step explanation:

Data given and notation  

[tex]\bar X[/tex] represent the mean height for the sample  

[tex]s[/tex] represent the sample standard deviation

[tex]n=23[/tex] sample size  

[tex]\mu_o =15[/tex] represent the value that we want to test

t would represent the statistic (variable of interest)  

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

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is higher than 15, the system of hypothesis would be:  

Null hypothesis:[tex]\mu \leq 15[/tex]  

Alternative hypothesis:[tex]\mu > 15[/tex]  

If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex]  (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

For this case the statistic is given:

[tex] t = 2.24[/tex]

P-value

The first step is calculate the degrees of freedom, on this case:  

[tex]df=n-1=23-1=22[/tex]  

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

[tex]p_v =P(t_{(22)}>2.24)=0.01776[/tex]  

And for this case we can conclude that:

[tex] 0.01 < p_v < 0.025[/tex]

And we will reject the null hypothesis at [tex] \alpha=0.025[/tex] since [tex] p_v < \alpha[/tex]

With a one-sample t-test (t = 2.24, n = 23) for H0: μ = 15 vs. Hα: μ > 15, P-value (0.01 < P < 0.025) supports rejecting the null at α = 0.025. The claim "P-value > 0.1" is incorrect.

Based on the given information, we have a one-sample t statistic from a sample of n = 23 observations for the two-sided test of H0: μ = 15 versus Hα: μ > 15, with a t value of 2.24.

To determine the P-value, we compare the t value to a t-distribution with n-1 degrees of freedom. In this case, since we have n = 23 observations, we would compare the t value of 2.24 to the t-distribution with 22 degrees of freedom.

The P-value is the probability of observing a t value as extreme as 2.24 or more extreme, assuming the null hypothesis is true.

Now, let's evaluate the given statements:

1. "0.01 < P-value < 0.025 and we would reject the null hypothesis at α = 0.025 are both correct."

Since the P-value is between 0.01 and 0.025, it falls within the critical region for α = 0.025. This means that the P-value is smaller than α, so we would reject the null hypothesis at α = 0.025. Therefore, the statement is correct.

2. "P-value > 0.1. We would reject the null hypothesis at α = 0.025."

Since the given P-value is not greater than 0.1, this statement is incorrect. If the P-value is larger than the significance level α (in this case, 0.025), we would fail to reject the null hypothesis. In other words, we do not have enough evidence to conclude that the population mean is greater than 15.

Based on the above analysis, the correct statement is:

- "0.01 < P-value < 0.025 and we would reject the null hypothesis at α = 0.025 are both correct."

It is important to note that the decision to reject or fail to reject the null hypothesis depends on the chosen significance level α and the P-value. The P-value measures the strength of the evidence against the null hypothesis, while the significance level determines the threshold for rejecting the null hypothesis.

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Answer:

[tex] y'' + \frac{7}{t} y = 1[/tex]

For this case we can use the theorem of Existence and uniqueness that says:

Let p(t) , q(t) and g(t) be continuous on [a,b] then the differential equation given by:

[tex] y''+ p(t) y' +q(t) y = g(t) , y(t_o) =y_o, y'(t_o) = y'_o[/tex]

has unique solution defined for all t in [a,b]

If we apply this to our equation we have that p(t) =0 and [tex] q(t) = \frac{7}{t}[/tex] and [tex] g(t) =1[/tex]

We see that [tex] q(t)[/tex] is not defined at t =0, so the largest interval containing 1 on which p,q and g are defined and continuous is given by [tex] (0, \infty)[/tex]

And by the theorem explained before we ensure the existence and uniqueness on this interval of a solution (unique) who satisfy the conditions required.

Step-by-step explanation:

For this case we have the following differential equation given:

[tex] t y'' + 7y = t[/tex]

With the conditions y(1)= 1 and y'(1) = 7

The frist step on this case is divide both sides of the differential equation by t and we got:

[tex] y'' + \frac{7}{t} y = 1[/tex]

For this case we can use the theorem of Existence and uniqueness that says:

Let p(t) , q(t) and g(t) be continuous on [a,b] then the differential equation given by:

[tex] y''+ p(t) y' +q(t) y = g(t) , y(t_o) =y_o, y'(t_o) = y'_o[/tex]

has unique solution defined for all t in [a,b]

If we apply this to our equation we have that p(t) =0 and [tex] q(t) = \frac{7}{t}[/tex] and [tex] g(t) =1[/tex]

We see that [tex] q(t)[/tex] is not defined at t =0, so the largest interval containing 1 on which p,q and g are defined and continuous is given by [tex] (0, \infty)[/tex]

And by the theorem explained before we ensure the existence and uniqueness on this interval of a solution (unique) who satisfy the conditions required.

Answer:

The longest interval in which the given initial value problem is certain to have a unique twice-differentiable solution is (0,∞)

Step-by-step explanation:

Given the differential equation:

ty'' + 7y = t .................................(1)

Together with the initial conditions:

y(1) = 1, y'(1) = 7

We want to determine the longest interval in which the given initial value problem is certain to have a unique twice-differentiable solution.

First, let us have the differential equation (1) in the form:

y'' + p(t)y' + q(t)y = r(t) ..................(2)

We do that by dividing (1) by t

So that

y''+ (7/t)y = 1 ....................................(3)

Comparing (3) with (2)

p(t) = 0

q(t) = 7/t

r(t) = 1

For t = 0, p(t) and r(t) are continuous, but q(t) = 7/0, which is undefined. Zero is certainly out of the required points.

In fact (-∞, 0) and (0,∞) are the points where p(t), q(t) and r(t) are continuous. But t = 1, which is contained in the initial conditions is found in (0,∞), and that makes it the correct interval.

So the largest interval containing 1 on which p(t), q(t) and r(t) are defined and continuous is (0,∞)

It appears that the question is incomplete but the answer to the part given is as below.

Answer:

(a) 0.0313

Step-by-step explanation:

In a normal distribution curve, the mean divides the curve into two equal parts. Hence, the probability of being lesser or higher than the mean is 1/2. This is the probability for a single day.

From the question, the probability of a day is independent of that of another day. For 5 days, the probability is

[tex](\frac{1}{2})^5 = 0.5^5 = 0.03125 = 0.0313[/tex] to 3 significant digits.

Answer:

a. Null hypothesis: The mean plasma aluminum level of the population of infants receiving antacids is equal to the mean plasma aluminum level of the population of infants not receiving antacids.

Complementary alternative hypothesis: The mean plasma aluminum level of the population of infants receiving antacids is different from the mean plasma aluminum level of the population of infants not receiving antacids.

b. (32.1, 42.3)

c. p-value < .00001

d. The null hypothesis is rejected at the α=0.05 significance level

e. Reformulated null hypothesis: The mean plasma aluminum level of the population of infants receiving antacids is equal to the mean plasma aluminum level of the population of infants not receiving antacids.

Reformulated complementary alternative hypothesis: The mean plasma aluminum level of the population of infants receiving antacids is higher than the mean plasma aluminum level of the population of infants not receiving antacids.

p-value equals < .00001. The null hypothesis is rejected at the α=0.05 significance level. This suggests that being given antacids greatly increases the plasma aluminum levels of children.

Step-by-step explanation:

a. Null hypothesis: The mean plasma aluminum level of the population of infants receiving antacids is equal to the mean plasma aluminum level of the population of infants not receiving antacids.

Complementary alternative hypothesis: The mean plasma aluminum level of the population of infants receiving antacids is different from the mean plasma aluminum level of the population of infants not receiving antacids. This may imply that being given antacids significantly changes the plasma aluminum level of infants.

b. Since the population standard deviation σ is unknown, we must use the t distribution to find 95% confidence limits for μ. For a t distribution with 10-1=9 degrees of freedom, 95% of the observations lie between -2.262 and 2.262. Therefore, replacing σ with s, a 95% confidence interval for the population mean μ is:

(X bar - 2.262\frac{s}{\sqrt{10} } , X bar + 2.262\frac{s}{\sqrt{10} })

Substituting in the values of X bar and s, the interval becomes:

(37.2 - 2.262\frac{7.13}{\sqrt{10} } , 37.2 + 2.262\frac{7.13}{\sqrt{10} })

or (32.1, 42.3)

c. To calculate p-value of the sample , we need to calculate the t-statistics which equals:

t=\frac{(Xbar-u)}{\frac{s}{\sqrt{10} } } = \frac{(37.2-4.13)}{\frac{7.13}{\sqrt{10} } } = 14.67.

Given two-sided test and degrees of freedom = 9, the p-value equals < .00001, which is less than 0.05.

d. The mean plasma aluminum level for the population of infants not receiving antacids is 4.13 ug/l - not a plausible value of mean plasma aluminum level for the population of infants receiving antacids. The 95% confidence interval for the population mean of infants receiving antacids is (32.1, 42.3) and does not cover the value 4.13. Therefore, the null hypothesis is rejected at the α=0.05 significance level. This suggests that being given antacids greatly changes the plasma aluminum levels of children.

e. Reformulated null hypothesis: The mean plasma aluminum level of the population of infants receiving antacids is equal to the mean plasma aluminum level of the population of infants not receiving antacids.

Reformulated complementary alternative hypothesis: The mean plasma aluminum level of the population of infants receiving antacids is higher than the mean plasma aluminum level of the population of infants not receiving antacids.

Given one-sided test and degree of freedom = 9, the p-value equals < .00001, which is less than 0.05. This result is similar to result in part (c). the null hypothesis is rejected at the α=0.05 significance level. This suggests that being given antacids greatly increases the plasma aluminum levels of children.

To test whether the mean plasma aluminum level of infants on antacids differs from those not on antacids, a null hypothesis (that the means are equal) is established alongside an alternative. A confidence interval and p-value are calculated to assess this hypothesis, and based on these results, a decision is made to reject or not reject the null hypothesis.

Hypotheses Formulation and Test Statistics

To analyze the plasma aluminum levels in infants receiving antacids compared to those not receiving them, one would perform a hypothesis test. The steps include formulating a null hypothesis (H0) and an alternative hypothesis (Ha), calculating the test statistic, finding the p-value, and making a decision regarding H0 based on the p-value and the confidence interval.

Null Hypothesis (H0): The mean plasma aluminum level of infants receiving antacids is equal to the mean level of those not receiving antacids (H0: μ = 4.13 µg/L).

Alternative Hypothesis (Ha): The mean plasma aluminum level of infants receiving antacids is not equal to the mean level of those not receiving antacids (Ha: μ ≠ 4.13 µg/L).

To construct a 95% confidence interval, we use the sample mean (× = 37.20 µg/L), sample standard deviation (s = 7.13 µg/L), the sample size (n = 10), and the t-distribution since the population variance is unknown. The confidence interval provides a range of values within which the true mean is likely to lie.

For the p-value, we compare it against the alpha level α=0.05. If the p-value is less than α, we reject H0; otherwise, we do not reject H0. The p-value indicates the likelihood of obtaining a sample mean at least as extreme as the one observed if H0 were true.

If the confidence interval does not include the population mean of children not receiving antacids and the p-value is less than α, we reject H0 in favor of Ha. If a one-sided test is appropriate (for example, if we only want to test if the mean aluminum level is higher in the treated group), Ha would be reformulated accordingly (Ha: μ > 4.13 µg/L), potentially resulting in a different decision from the two-sided test.

Answer:

Step-by-step explanation:

We need to observe total 9 days.First we need to choose any 4 days from the 9 days. From these total 9 days, 4 days can be chosen in [tex]^9C_4 = \frac{9!}{4!\times5!} = \frac{6\times7\times8\times9}{4!} = 126[/tex] ways.

The probability of occurring an auto related accident is [tex]\frac{52}{100}[/tex].

Similarly, the probability of not occurring an accident is [tex]\frac{100 - 52}{100} = \frac{48}{100}[/tex].

Hence, the required probability is [tex]126\times(\frac{52}{100} )^4(\frac{48}{100} )^5[/tex].

Answer:

0.1222 or 12.22%

Yes, it is unlikely.

Step-by-step explanation:

If the probability of someone being infected is 0.005, then the probability of someone not being infected is 0.995. In order for the combined sample to test negative (N), all of the 26 people must test negative. Thus, the probability that the combined sample tests positive is:

[tex]P(Sample = P) = 1 -P(N=26)\\P(Sample = P) = 1 -0.995^{26}\\P(Sample = P) = 0.1222=12.22\%[/tex]

There is a 0.1222 or 12.22% probability that the combined sample tests positive, which is unlikely to occur.

Final answer:

To calculate the magnitude of the torque exerted on a bolt, use the formula τ = rFsinθ with the given values: lever arm length of 6 inches, force of 16 lbs, and an assumed angle of 90 degrees. The calculated torque is 96 in.lb

Explanation:

The magnitude of torque (τ) can be calculated using the formula τ = rFsinθ, where r is the distance from the pivot to the point where the force is applied, F is the force applied, and θ is the angle between the force and the arm of the lever—in this case, assuming the force is perpendicular (90 degrees) to the lever arm for maximum torque.

Given that |vector PQ| = 6 in. as the lever arm (r) and |F| = 16 lb. as the force applied, and assuming the angle θ is 90 degrees (since it's not provided but necessary for calculating maximum torque):

τ = rFsinθ = (6 in.)(16 lb.)(sin90°) = 96 in.lb. because sin90° = 1.

Therefore, the magnitude of the torque exerted by the force on the bolt at point P is 96 in.lb.

Options: A. Chi squared Statistics

B. ANOVA

C. Independent samples t-test.

D. z-test of a population proportion.

Answer:A. Chi squared Statistics

Step-by-step explanation: A Chi Squared Statistics is a Statistical technique or test measure that determines how an expectation compares to actual observation. Chi squared Statistics data is expected to have the following features

Such as the data must be RAW, RANDOM, MUTUALLY EXCLUSIVE, OBTAINED FROM INDEPENDENT VARIABLES, AND LARGE SAMPLES WHICH WILL BE ENOUGH.

the survey of one hundred adults from each ethnic/racial groups means the data possess all the characters of a Chi Squared Statistics or test measure.

Answer:

a) Bi [P ( X >=15 ) ] ≈ 0.9944

b) Bi [P ( X >=30 ) ] ≈ 0.3182

c)  Bi [P ( 25=< X =< 35 ) ] ≈ 0.6623

d) Bi [P ( X >40 ) ] ≈ 0.0046  

Step-by-step explanation:

Given:

- Total sample size n = 745

- The probability of success p = 0.037

- The probability of failure q = 0.963

Find:

a. 15 or more will live beyond their 90th birthday

b. 30 or more will live beyond their 90th birthday

c. between 25 and 35 will live beyond their 90th birthday

d. more than 40 will live beyond their 90th birthday

Solution:

- The condition for normal approximation to binomial distribution:                                                

                    n*p = 745*0.037 = 27.565 > 5

                    n*q = 745*0.963 = 717.435 > 5

                    Normal Approximation is valid.

a) P ( X >= 15 ) ?

 - Apply continuity correction for normal approximation:

                Bi [P ( X >=15 ) ] = N [ P ( X >= 14.5 ) ]

 - Then the parameters u mean and σ standard deviation for normal distribution are:

                u = n*p = 27.565

                σ = sqrt ( n*p*q ) = sqrt ( 745*0.037*0.963 ) = 5.1522

- The random variable has approximated normal distribution as follows:

                X~N ( 27.565 , 5.1522^2 )

- Now compute the Z - value for the corrected limit:

                N [ P ( X >= 14.5 ) ] = P ( Z >= (14.5 - 27.565) / 5.1522 )

                N [ P ( X >= 14.5 ) ] = P ( Z >= -2.5358 )

- Now use the Z-score table to evaluate the probability:

                P ( Z >= -2.5358 ) = 0.9944

                N [ P ( X >= 14.5 ) ] = P ( Z >= -2.5358 ) = 0.9944

Hence,

                Bi [P ( X >=15 ) ] ≈ 0.9944

b) P ( X >= 30 ) ?

 - Apply continuity correction for normal approximation:

                Bi [P ( X >=30 ) ] = N [ P ( X >= 29.5 ) ]

- Now compute the Z - value for the corrected limit:

                N [ P ( X >= 29.5 ) ] = P ( Z >= (29.5 - 27.565) / 5.1522 )

                N [ P ( X >= 29.5 ) ] = P ( Z >= 0.37556 )

- Now use the Z-score table to evaluate the probability:

                P ( Z >= 0.37556 ) = 0.3182

                N [ P ( X >= 29.5 ) ] = P ( Z >= 0.37556 ) = 0.3182

Hence,

                Bi [P ( X >=30 ) ] ≈ 0.3182  

c) P ( 25=< X =< 35 ) ?

 - Apply continuity correction for normal approximation:

                Bi [P ( 25=< X =< 35 ) ] = N [ P ( 24.5=< X =< 35.5 ) ]

- Now compute the Z - value for the corrected limit:

                N [ P ( 24.5=< X =< 35.5 ) ]= P ( (24.5 - 27.565) / 5.1522 =<Z =< (35.5 - 27.565) / 5.1522 )

                N [ P ( 24.5=< X =< 25.5 ) ] = P ( -0.59489 =<Z =< 1.54011 )

- Now use the Z-score table to evaluate the probability:

                P ( -0.59489 =<Z =< 1.54011 ) = 0.6623

               N [ P ( 24.5=< X =< 35.5 ) ]= P ( -0.59489 =<Z =< 1.54011 ) = 0.6623

Hence,

                Bi [P ( 25=< X =< 35 ) ] ≈ 0.6623

d) P ( X > 40 ) ?

 - Apply continuity correction for normal approximation:

                Bi [P ( X >40 ) ] = N [ P ( X > 41 ) ]

- Now compute the Z - value for the corrected limit:

                N [ P ( X > 41 ) ] = P ( Z > (41 - 27.565) / 5.1522 )

                N [ P ( X > 41 ) ] = P ( Z > 2.60762 )

- Now use the Z-score table to evaluate the probability:

               P ( Z > 2.60762 ) = 0.0046

               N [ P ( X > 41 ) ] =  P ( Z > 2.60762 ) = 0.0046

Hence,

                Bi [P ( X >40 ) ] ≈ 0.0046  

Using the normal approximation to binomial is appropriate for the given problem. To find probabilities, we calculate the mean and standard deviation of the binomial, then calculate the z-scores and use the normal distribution to estimate the required probabilities for each case.

To determine if it is appropriate to use the normal approximation to the binomial distribution, we must confirm that both np and n(1-p) are greater than 5, where n is the number of trials and p is the probability of success. For a class of 745 students and probability of living past 90 being 3.7%, we have:

Since both values are greater than 5, we can proceed with the normal approximation.

Calculations using Normal Approximation:

a. To find the probability of 15 or more students living past 90, we calculate the mean (μ) and standard deviation (σ) for the binomial distribution:

Next, we find the z-score for 14.5 (since we need more than 15, we use the continuity correction factor of 0.5) and use the standard normal distribution to estimate the probability:

P(X ≥ 15) = 1 - P(X < 15) = 1 - P(Z < (14.5 - μ) / σ)

b. - d. The procedure is similar for parts b, c, and d: calculate the z-scores for the respective values using the binomial mean and standard deviation, and then use the normal distribution to find the probabilities. Based on these calculations, we can provide the required estimates.

Answer:

Step-by-step explanation:

Hello!

The population of the study is the Toco Toucan, the largest member of the toucan family. It is believed that the length and size of the beak are due to its function of dissipating heat (a cooling mechanism).

You have two variables of interest.

X: Outdoor temperature.

Temperature (ºC) 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

Y: Total beak heat loss, as a percent of the total body heat loss of a toucan.

Percent heat loss from beak 33 34 33 36 36 47 52 51 41 50 49 50 55 60 60 62

The objective is to construct a least-squares regression line to predict the body heat loss of the toucans given the outdoor temperature.

Using a statistical software I estimated the regression model:

^Yi= 2.71 + 1.96Xi

1. To predict what will be the value of the beak heat loss as a percent of the total body heat loss at a temperature of 25ºC you have to simply replace the value of X in the equation:

^Yi= 2.71 + 1.96 (25)= 51.71ºC

The expected beak heat loss given an outdoor temperature of 25ºC is 51.71ºC.

2. To know what percent of the variation of the beak heat loss is explained by the outdoor temperature you have to calculate the coefficient of determination.

R²= 0.85

This means that 85% of the variability of the beak heat loss as a percent of the total body heat loss of the Toco Toucans is explained by the outdoor Temperature under the estimated model ^Yi= 2.71 + 1.96Xi

3. The correlation coefficient between these two variables is r= 0.92, this means that there is a strong positive linear correlation between the beak heat loss and the outdoor temperature. This means that when the outdoor temperature rises, the beak heat loss increases.

I hope it helps!

Answer:

The critical value is 6.314

Step-by-step explanation:

Null hypothesis: The proportion of businesses that earn a profit within the first two years of operation is 80%

Alternate hypothesis: The proportion of businesses e earn a profit within the first two years of operation is greater than 18%

The hypothesis test has one critical value because it is a one-tailed test. It is a one-tailed test because the alternate hypothesis is expressed using the inequality, greater than.

n = 2

degree of freedom = n - 1 = 2 - 1 = 1

significance level = 10%

Using the t-distribution table, critical value corresponding to 1 degree of freedom and 10% significance level is 6.314.

Answer:

To determine the critical value or values for a one-proportion z-test at the 10% significance level when the hypothesis test is left- or right-tailed, we must use the look-up table for zz0.01. Since this is a right-tailed test, our critical value is positive 1.282.

Answer:

The calculated 99% confidence interval is wider than the 95% confidence interval.      

Step-by-step explanation:

We are given the following in the question:

95% confidence interval for the population proportion

(0.65, 0.69)

Let [tex]\hat{p}[/tex] be the sample proportion

Confidence interval:

[tex]p \pm z_{stat}(\text{Standard error})[/tex]

[tex]z_{critical}\text{ at}~\alpha_{0.05} = 1.96[/tex]

Let x be the standard error, then, we can write

[tex]\hat{p} - 1.96x = 0.65\\\hat{p}+1.96x = 0.69[/tex]

Solving the two equations, we get,

[tex]2\hat{p} = 0.65 + 0.69\\\\\hat{p} = \dfrac{1.34}{2} = 0.67\\\\x = \dfrac{0.69 - 0.67}{1.96} \approx 0.01[/tex]

99% Confidence interval:

[tex]p \pm z_{stat}(\text{Standard error})[/tex]

[tex]z_{critical}\text{ at}~\alpha_{0.01} = 2.58[/tex]

Putting values, we get,

[tex]0.67 \pm 2.58(0.01)\\=0.67 \pm 0.0258\\=(0.6442,0.6958)[/tex]

Thus, the calculated 99% confidence interval is wider than the 95% confidence interval .

Answer:

(A) The probability that a randomly selected adult is either overweight or obese is 0.688.

(B) The probability that a randomly selected adult is neither overweight nor obese is 0.312.

(C) The events "overweight" and "obese" exhaustive.

(D) The events "overweight" and "obese" mutually exclusive.

Step-by-step explanation:

Denote the events as follows:

X = a person is overweight

Y = a person is obese.

The information provided is:

A person is overweight if they have BMI 25 or more but below 30.

A person is obese if they have BMI 30 or more.

P (X) = 0.331

P (Y) = 0.357

(A)

The events of a person being overweight or obese cannot occur together.

Since if a person is overweight they have (25 ≤ BMI < 30) and if they are obese they have BMI ≥ 30.

So, P (X ∩ Y) = 0.

Compute the probability that a randomly selected adult is either overweight or obese as follows:

[tex]P(X\cup Y)=P(X)+P(Y)-P(X\cap Y)\\=0.331+0.357-0\\=0.688[/tex]

Thus, the probability that a randomly selected adult is either overweight or obese is 0.688.

(B)

Commute the probability that a randomly selected adult is neither overweight nor obese as follows:

[tex]P(X^{c}\cup Y^{c})=1-P(X\cup Y)\\=1-0.688\\=0.312[/tex]

Thus, the probability that a randomly selected adult is neither overweight nor obese is 0.312.

(C)

If two events cannot occur together, but they form a sample space when combined are known as exhaustive events.

For example, flip of coin. On a flip of a coin, the flip turns as either Heads or Tails but never both. But together the event of getting a Heads and Tails form a sample space of a single flip of a coin.

In this case also, together the event of a person being overweight or obese forms a sample space of people who are heavier in general.

Thus, the events "overweight" and "obese" exhaustive.

(D)

Mutually exclusive events are those events that cannot occur at the same time.

The events of a person being overweight and obese are mutually exclusive.

Answer:

21 minutes

Step-by-step explanation:

Her expected commute time is given by the probability of having to wait for the train (0.2) multiplied by the commute time in this scenario (25 min), added to the probability of not having to wait for the train (1 - 0.2) multiplied by the commute time in this scenario (20 min). The expected commute time is:

[tex]E(X) = 0.2*25+(1-0.2)*20\\E(X) = 21\ minutes[/tex]

Her expected commute time is 21 minutes.

Answer:

(a) Average number of cars in the system is 1

(b) Average time a car waits is 12 minutes

(c) Average time a car spends in the system is 2 minutes

(d) Probability that an arrival has to wait for service is 0.08.

Step-by-step explanation:

We are given the following

Arrival Rate, A = 2.5

Service Rate B = 5

(a) Average Number of Cars in the System is determined by dividing the Arrival Rate A by the difference between the Service Rate B, and Arrival Rate A.

Average number of cars = A/(B - A)

= 2.5/(5 - 2.5)

= 2.5/2.5 = 1

There is an average of 1 car.

(b) Average time a car waits = A/B(B - A)

= 2.5/5(5 - 2.5)

= 2.5/(5 × 2.5)

= 2.5/12.5

= 1/5

= 0.20 hours

Which is 12 minutes

(c) Average time a car spends in the system is the ratio of the average time a car waits to the service rate.

Average time = 0.2/5

= 0.04 hours

= 2.4 minutes

Which is approximately 2 minutes.

(d) Probability that an arrival has to wait for service is the ratio of the average time a car waits to rate of arrivals.

Probability = 0.2/2.5

= 0.08

The average number of cars in the system can be found using Little's Law. The average time a car waits for the oil and lubrication service to begin is half of the average time in the system. The probability that an arrival has to wait for service is obtained by dividing the arrival rate by the service rate.

(a) To find the average number of cars in the system, we can use Little's Law. Little's Law states that the average number of cars in the system (L) equals the arrival rate (λ) multiplied by the average time a car spends in the system (W). In this case, λ = 2.5 cars per hour and the service rate (μ) = 5 cars per hour.

So, L = λ * W. Rearranging the formula, W = L / λ. Substituting the values, W = (2.5 / 5) = 0.5 hours or 30 minutes.

(b) The average time that a car waits for the oil and lubrication service to begin is equal to half of the average time a car spends in the system, which is 30 minutes.

(c) The average time a car spends in the system is equal to the average waiting time plus the average service time. The average service time (1/μ) in this case is 1/5 hour or 12 minutes. Therefore, the average time a car spends in the system is 30 minutes + 12 minutes = 42 minutes.

(d) To find the probability that an arrival has to wait for service, we can use the formula P(wait) = λ / μ = 2.5 / 5 = 0.5 or 50%

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Answer:

[tex]\sigma^2=73.96 \ kg^2[/tex]

Step-by-step explanation:

Standard Deviation and Variance

If we have a data set of measured values the variance is defined as the average of the squared differences that each value has from the mean. The formula to calculate the variance is:

[tex]\displaystyle \sigma^2=\frac{\sum(x_i-\mu)^2}{n}[/tex]

Where [tex]\mu[/tex] is the mean of the measured values xi (i running from 1 to n), and n is the total number of values.

[tex]\displaystyle \mu=\frac{\sum x_i}{n}[/tex]

The standard deviation is known by the symbol [tex]\sigma[/tex] and is the square root of the variance. We know the standar deviation of the weight in kg of a group of teenagers to be 8.6 kg. Thus, the variance is

[tex]\sigma^2=8.6^2=73.96 \ kg^2[/tex]

[tex]\boxed{\sigma^2=73.96 \ kg^2}[/tex]

Answer:

A

Step-by-step explanation:

Answer:

See below

Step-by-step explanation:

Recall first:

Given a vector  

[tex](x_1,x_2,...,x_n)[/tex]

Euclidean norm

[tex]\sqrt{x_1^2+x_2^2+...+x_n^2}[/tex]

Sum norm

[tex]|x_1|+|x_2|+...+|x_n|[/tex]

Max norm

[tex]Max\{|x_1|,|x_2|,...|x_n|\}[/tex]

Now let us apply these definitions to our vectors

Vector (1,1,1)

Euclidean norm

[tex]\sqrt{1^2+1^2+1^2}=\sqrt{3}[/tex]

Sum norm

|1|+|1|+|1| = 3

Max norm

Max{|1|, |1|, |1|} = |1| = 1

Vector (3,0,0)

Euclidean norm

[tex]\sqrt{3^2+0^2+0^2}=\sqrt{3^2}=3[/tex]

Sum norm

|3|+|0|+|0| = 3

Max norm

Max{|3|, |0|, |0|} = |3| = 3

Vector (-1,1,4)

Euclidean norm

[tex]\sqrt{(-1)^2+1^2+4^2}=\sqrt{18}[/tex]

Sum norm

|-1|+|1|+|4| = 1+1+4 =6

Max norm

Max{|-1|, |1|, |4|} = |4| = 4

Vector (-1.4, 3)

Euclidean norm

[tex]\sqrt{(-1.4)^2+3^2}=\sqrt{10.96}[/tex]

Sum norm

|-1.4|+|3| = 1.4+3 = 4.4

Max norm

Max{|-1.4|, |3|} = |3| = 3

Vector (4,4,4,4)

Euclidean norm

[tex]\sqrt{4^2+4^2+4^2+4^2}=\sqrt{4*4^2}=8[/tex]

Sum norm

|4|+|4|+|4|+|4| = 16

Max norm

Max{|4|, |4|, |4|, |4|} = |4| = 4

Final Answer:

- Vector (a) has a Euclidean norm of approximately 1.732, a sum norm of 3, and a max norm of 1.
- Vector (b) has a Euclidean norm of 3, a sum norm of 3, and a max norm of 3.
- Vector (c) has a Euclidean norm of approximately 4.243, a sum norm of 6, and a max norm of 4.
- Vector (d) has a Euclidean norm of approximately 4.123, a sum norm of 5, and a max norm of 4.
- Vector (e) has a Euclidean norm of 8, a sum norm of 16, and a max norm of 4.

Explanation:

Sure, let's calculate the norms for each of the given vectors:

(a) For the vector (1, 1, 1):
- The Euclidean norm (also known as the L2 norm) is the square root of the sum of the squares of the components. For this vector, it's [tex]\(\sqrt{1^2 + 1^2 + 1^2} = \sqrt{3} \approx 1.732\)[/tex].
- The sum norm (or L1 norm) is the sum of the absolute values of the components. For this vector, it's |1| + |1| + |1| = 3.
- The max norm (also known as the infinity norm) is the maximum absolute value component of the vector. For this vector, it's [tex]\(\max(|1|, |1|, |1|) = 1\)[/tex].

(b) For the vector (3, 0, 0):
- The Euclidean norm of this vector is [tex]\(\sqrt{3^2 + 0^2 + 0^2} = \sqrt{9} = 3\)[/tex].
- The sum norm of this vector is |3| + |0| + |0| = 3.
- The max norm of this vector is max(|3|, |0|, |0|) = 3.

(c) For the vector (-1, 1, 4):
- The Euclidean norm of this vector is [tex]\(\sqrt{(-1)^2 + 1^2 + 4^2} = \sqrt{1 + 1 + 16} = \sqrt{18} \approx 4.243\)[/tex].
- The sum norm is (|-1| + |1| + |4| = 1 + 1 + 4 = 6).
- The max norm is max(|-1|, |1|, |4|) = 4.

(d) For the vector (-1, 4):
- The Euclidean norm is [tex]\(\sqrt{(-1)^2 + 4^2} = \sqrt{1 + 16} = \sqrt{17} \approx 4.123\)[/tex].
- The sum norm is |-1| + |4| = 1 + 4 = 5.
- The max norm is max(|-1|, |4|) = 4.

(e) For the vector (4, 4, 4, 4):
- The Euclidean norm is [tex]\(\sqrt{4^2 + 4^2 + 4^2 + 4^2} = \sqrt{16 + 16 + 16 + 16} = \sqrt{64} = 8\)[/tex].
- The sum norm is |4| + |4| + |4| + |4| = 4 + 4 + 4 + 4 = 16.
- The max norm is max(|4|, |4|, |4|, |4|) = 4.

Now we have calculated the three types of norms for each of the vectors:
- Vector (a) has a Euclidean norm of approximately 1.732, a sum norm of 3, and a max norm of 1.
- Vector (b) has a Euclidean norm of 3, a sum norm of 3, and a max norm of 3.
- Vector (c) has a Euclidean norm of approximately 4.243, a sum norm of 6, and a max norm of 4.
- Vector (d) has a Euclidean norm of approximately 4.123, a sum norm of 5, and a max norm of 4.
- Vector (e) has a Euclidean norm of 8, a sum norm of 16, and a max norm of 4.

Answer:

C. 0.896

Step-by-step explanation:

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

Binomial probability distribution

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

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

The expected value of the binomial distribution is:

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

The standard deviation of the binomial distribution is:

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

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

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

In this problem, we have that:

[tex]n = 80, p = 0.76[/tex]

So

[tex]E(X) = np = 80*0.76 = 60.8[/tex]

[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{80*0.76*0.24} = 3.82[/tex]

What is the probability that at least seventy percent of the sample prefers Coke to Pepsi?

0.7*80 = 56.

This probability is 1 subtracted by the pvalue of Z when X = 56. So

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

[tex]Z = \frac{56 - 60.8}{3.82}[/tex]

[tex]Z = -1.26[/tex]

[tex]Z = -1.26[/tex] has a pvalue of 0.1040.

1 - 0.1040 = 0.8960

So the correct answer is:

C. 0.896

Answer:

The typical error between a predicted college GPA using this regression model and an actual college GPA for a given student will be about 0.374 grade points in size (absolute value).

Step-by-step explanation:

The linear regression line for College GPA based on High school GPA is:

College GPA = 1.07 + 0.62 High-school GPA

It is provided that the standard error of the regression line is,

[tex]s_{e}=0.374[/tex]

The standard error of a regression line is the average distance between the predicted value and the regression equation.

It is the square root of the average squared deviations.

It is also known as the standard error of estimate.

The standard error of 0.374 implies that:

The typical error between a predicted college GPA using this regression model and an actual college GPA for a given student will be about 0.374 grade points in size (absolute value).

The standard error of the regression tells us the average amount that the actual college GPA deviates from the predicted college GPA. A smaller standard error indicates a better fit of the model to the data. In this case, the small standard error suggests that the linear regression model provides a good prediction of college GPA based on high school GPA for the students at the local college.

The standard error of the regression tells us the average amount that the actual college GPA deviates from the predicted college GPA based on the high school GPA. In this case, the standard error of the regression is 0.374. This means that, on average, the actual college GPA for a student deviates from the predicted college GPA by approximately 0.374.

This value gives us an idea of the accuracy of the linear regression model in predicting college GPA based on high school GPA. A smaller standard error indicates a better fit of the model to the data, implying that the predicted college GPA is closer to the actual college GPA. Conversely, a larger standard error suggests that the model's predictions are less accurate.

In this case, the standard error of the regression is relatively small (0.374), which indicates that the linear regression model provides a good prediction of college GPA based on high school GPA for the students at the local college.

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Final answer:

The experimental unit in the given scenario is the pine tree seedlings which are used to test the hypothesis regarding their growth in shade and adaptation to drought conditions.

Explanation:

The experimental unit or subject in the described experiment is b) pine tree seedlings. These seedlings are what the investigators manipulate (by altering light conditions) and measure (by weighing after drying) to test the hypothesis regarding the growth of pines in shade as an adaptation to resist drought.

This experiment tests how pine seedlings grow in different lighting conditions to characterize features like growth rate and drought resistance, informed by observations such as acclimatization where the structure of leaves change when transitioning from sun to shade or vice versa to achieve photosynthetic efficiency.

The true statements are:

Let's analyze each statement based on the provided stem-and-leaf plot:

Given stem-and-leaf plot:

Stem | Leaf

 0  | 0 1 4 5 5 8

 1  | 0 0 4 5 8

a. The median annual wage estimate was $85,000 per year for these 15 architecture and engineering occupations.

The median is the middle value when the data is arranged in ascending order.

The middle value is the 8th value, which is 85 (thousands of dollars).

This statement is true.

b. The modal class is 100 to 110 thousand dollars, with five occupations having salaries in that class.

The modal class is the class with the highest frequency.

In this case, the class 100 to 110 thousand dollars has three frequencies. This statement is false.

c. The shape of the distribution is skewed to the right.

Skewed to the right means that the distribution's tail is on the right side. Looking at the stem-and-leaf plot, we can see that the data has more values on the left side and tails off to the right.

This statement is true.

d. The modal annual wage estimate was $92,000 per year for these 15 architecture and engineering occupations.

The mode is the most frequently occurring value.

The mode in this data set is not explicitly shown, but we can see that there is no class with a significantly higher frequency.

This statement is false.

e. The annual wage of $108,000 was the largest wage estimate for architecture and engineering occupations in May 2014.

The largest value in the data set is 108 (thousands of dollars), which corresponds to an annual wage of $108,000.

This statement is true.

So, the true statements are:

a. The median annual wage estimate was $85,000 per year for these 15 architecture and engineering occupations.

c. The shape of the distribution is skewed to the right.

e. The annual wage of $108,000 was the largest wage estimate for architecture and engineering occupations in May 2014.

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Without the visual plot, it's impossible to confirm any statements with certainty. However, a theoretical ordering of the given data suggests that the median annual wage isn't $85,000 but closer to $84,000, the claimed modal class has fewer than five salaries, and the shape does seem to be skewed right. The highest wage seems to be $158,000 not $108,000.

This question involves understanding and interpreting a stem-and-leaf plot. Unfortunately, without being able to view the plot, it's impossible to conclusively confirm any of the statements provided.

However, we can do a generalized analysis based on the given data. Typically, a stem-and-leaf plot represents numerical data in an order, which makes it easier to see certain statistics like median, mode and range, and also the distribution shape. If the statement that the stem represents 'tens' and leaf represents 'ones' is true, then the data arranged in ascending order would be: 50, 57, 60, 75, 75, 80, 84, 85, 100, 101, 104, 105, 108, 145, 158 (all in thousands).

From these, it's clear: a) The median annual wage is not $85,000, it's actually $84,000. b) There aren't five salaries in the 100-110k range, there are only four. So the modal class isn't $100,000 to $110,000. c) The distribution appears to be skewed to the right, given that higher wages are less common. d) The modal wage estimate is unclear without knowing how many times each wage appears in the original data. e) The highest wage is $158,000 (not $108,00), assuming that the stem-and-leaf plot does not have repeating values.

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Answer:

Yes, Hypothesis test represents a statistically significant difference from 50% .

Step-by-step explanation:

We are given that 400 students were randomly sampled from a large university, and 289 said they did not get enough sleep.

Let Null Hypothesis, [tex]H_0[/tex] : p = 0.50 {means from the students sampled from the university, proportion of them who did not get enough sleep is 50%}

Alternate Hypothesis, [tex]H_1[/tex] : p [tex]\neq[/tex] 0.50 {means from the students sampled from the university, proportion of them who did not get enough sleep is different from 50%}

The test statistics we will use here is;

        T.S. = [tex]\frac{\hat p - p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ~ N(0,1)

where, [tex]\hat p[/tex] = sample proportion = 289/400 = 0.7225

            n = sample of students = 400

So, test statistics = [tex]\frac{0.7225 - 0.50}{\sqrt{\frac{0.7225(1-0.7225)}{400} } }[/tex] = 3.143

Now, at 1% level of significance, z table gives critical value of 2.5758. Since our test statistics is more than the critical value which means test statistics will lie in the rejection region, so we have sufficient evidence to reject null hypothesis.

Therefore, we conclude that from the students sampled from the university, proportion of them who did not get enough sleep is different from 50%.

The question asks to perform a hypothesis test to determine if there is a significant difference between the proportion of university students who do not get enough sleep and the assumed 50%. The steps include stating hypotheses, calculating the test statistic, and comparing it with a critical value at a 0.01 significance level. A decision to reject or not reject the null hypothesis will be made based on this comparison.

The student's question involves conducting a hypothesis test to determine if there is a statistically significant difference between the proportion of university students who do not get enough sleep and the assumed proportion of 50%. To perform this test at the given significance level of 0.01, the following steps need to be taken:

For our example, with 289 out of 400 students not getting enough sleep, the sample proportion p-hat is 289/400 = 0.7225. Using the formula, the test statistic is then calculated and compared with the critical z-value for a 0.01 significance level. If the absolute value of the test statistic is greater than the critical z-value, the null hypothesis is rejected, suggesting that there is a statistically significant difference from 50%.

If the test statistic does not exceed the critical value, we fail to reject the null hypothesis, indicating that there is not enough evidence to suggest a significant deviation from 50%. Based on the results of the hypothesis test, conclusions about the sleep habits of university students can be drawn.

Answer:

The 95% confidence interval for the true proportion of defective batteries is (0.0966, 0.1034).

It is better to take a larger sample to derive conclusion about the true parameter value.

Step-by-step explanation:

The (1 - α) % confidence interval for proportion is:

[tex]CI=\hat p\pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]

Given:

n = 2000

X = 200

The sample proportion is:

[tex]\hat p=\frac{X}{n}=\frac{200}{2000}=0.10[/tex]

The critical value of z for 95% confidence interval is:

[tex]z_{\alpha /2}=z_{0.05/2}=z_{0.025}=1.96[/tex]

Compute the 95% confidence interval as follows:

[tex]CI=\hat p\pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}\\=0.10\pm1.96\times\sqrt{\frac{0.10(1-0.10)}{2000}}\\=0.10\pm0.0034\\=(0.0966, 0.1034)[/tex]

Thus, the 95% confidence interval for the true proportion of defective batteries is (0.0966, 0.1034).

Now if in a sample of 100 batteries there are 15 defectives, the the 95% confidence interval for this sample is:

[tex]CI=\hat p\pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}\\=0.15\pm1.96\times\sqrt{\frac{0.15(1-0.15)}{100}}\\=0.15\pm0.0706\\=(0.0794, 0.2206)[/tex]

It can be observed that as the sample size was decreased the width of the confidence interval was increased.

Thus, it can be concluded that it is better to take a larger sample to derive conclusion about the true parameter value.

Answer:

The probability of event A is 0.2160.

The probability of event B is 0.2153.

Step-by-step explanation:

Assume that the random variable X is defined as the number of defective components in a lot.

It is provided that of the 1060 component 229 are defective.

The probability of selecting a defective component is:

[tex]P(X)=\frac{229}{1060}=0.2160[/tex]

The proportion of defective components in a lot of 1060 is 0.2160.

It is provided that two components are selected to be tested.

Assuming the selection were without replacement.

A = the first component drawn is defective

B = the second component drawn is defective

        The probability of selecting a defective component from the entire lot

        of 1060 component is 0.2160.

        Thus, the probability of event A is 0.2160.

        According to event A, the first component selected was defective.

        So now there are 228 defective components among 1059  

        components.

         [tex]P(B)=\frac{228}{1059}= 0.2153[/tex]

        Thus, the probability of event B is 0.2153.

Both the probabilities are almost same.

This implies that the probability of selecting a defective component from the entire population of these components is approximately 0.2160.

Answer:

Required Probability = 0.0467

Step-by-step explanation:

We are given that a lot of 1060 components contains 229 that are defective.

Two components are drawn at random and tested.

A = event that the first component drawn is defective

B = event that the second component drawn is defective

So,P(first component drawn is defective,A)=(No. of defective component)÷

                                                                        (Total components)

P(A) = 229/1060

Similarly, P(second component drawn is defective,A) = 229/1060

Therefore, P(both component drawn defective) = [tex]\frac{229}{1060} * \frac{229}{1060}[/tex] = 0.0467 .

Answer:

2,075,673,600 batting orders may occur.

Step-by-step explanation:

The order of the first eight batters in the batting order is important. For example, if we exchange Jonathan Schoop with Adam Jones in the lineup, that is a different lineup. So we use the permutations formula to solve this problem.

Permutations formula:

The number of possible permutations of x elements from a set of n elements is given by the following formula:

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

First 8 batters

8 players from a set of 16. So

[tex]P_{(16,8)} = \frac{16!}{(16 - 8)!} = 518918400[/tex]

Last batter:

Any of the four pitchers.

How many different batting orders may​ occur?

4*518918400 = 2,075,673,600

2,075,673,600 batting orders may occur.

Final answer:

20,922,789,888,000 different ways.

Explanation:

To find out the number of different batting orders that may occur with the given constraints, we consider that there are 16 players who can play any position other than pitcher, and since the pitcher bats last, we only need to determine the batting order for the first 8 positions which any of the 16 players can fill. Since the order matters, we use permutations to calculate the number of ways to arrange these players. After the 8 positions are filled, the pitcher is automatically assigned the last batting slot.

The formula for permutations is:

P(n, k) = n! / (n-k)!,

where n is the total number of items to choose from, k is the number of items to arrange, and '!' denotes factorial which is the product of an integer and all the integers below it.

Therefore, the number of different batting orders is P(16, 8) which equals:

16! / (16-8)! = 16! / 8! = 20,922,789,888,000 possible ways.

This includes 1 of the 4 pitchers, who will be the last to bat in each permutation.

Answer:

0.3075 = 30.75% probability that a person will wait for more than 7 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.

The standard deviation is the square root of the variance.

In this problem, we have that:

[tex]\mu = 6, \sigma = \sqrt{4} = 2[/tex]

Find the probability that a person will wait for more than 7 minutes.

This is 1 subtracted by the pvalue of Z when X = 7. So

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

[tex]Z = \frac{7 - 6}{2}[/tex]

[tex]Z = 0.5[/tex]

[tex]Z = 0.5[/tex] has a pvalue of 0.6915

1 - 0.6915 = 0.3075

0.3075 = 30.75% probability that a person will wait for more than 7 minutes.

The probability that a person will wait for more than 7 minutes is 30.85%.

Z score is used to determine by how many standard deviations the raw score is above or below the mean. It is given by:

z = (raw score - mean) / standard deviation

Given;  mean of 6 minutes and variance = 4 minutes, hence:

Standard deviation = √variance = √4 = 2 minutes

For > 7 minutes:

z = (7 - 6)/2 = 0.5

P(z > 0.5) = 1 - P(z < 0.5) = 1 - 0.6915 = 0.3085

The probability that a person will wait for more than 7 minutes is 30.85%.

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The rabbit population doubles every 6 weeks. After converting 112 days to 16 weeks, we calculate the growth over 2.67 six-week periods to find that there will be approximately 61 rabbits after 112 days.

The rabbits population grows exponentially and doubles every 6 weeks. To determine the population after a certain time, we would typically use a formula like P(t) = P0 (2^(t/T)), where P(t) is the population at time t, P0 is the initial population, 2 represents the doubling factor, and T is the time period for doubling in the same units as t. However, we need to convert days to weeks since the doubling time is given in weeks. There are 7 days in a week, so 112 days is equal to 112/7 = 16 weeks.

Following this, we calculate the number of periods in 16 weeks by dividing 16 weeks by 6 (the number of weeks it takes for the population to double). This gives us 16/6 = approximately 2.67. The population after 16 weeks would be P(16) = 9 (2^(2.67)). Simplifying with a calculator, we find the population to be about 61 rabbits, rounding to the nearest whole number.

Answer:

(a) = 40%

(b) = 28%

(c) Expected value = $222,500

Standard deviation = $7,216.88

Step-by-step explanation:

This is a normal distribution with a = 210,000 and b =235,000

(a) The probability that he will get at least $225,000 for the house is:

[tex]P(X\geq 225,000) =1 -\frac{225,000-a}{b-a} =1-\frac{225,000-210,000}{235,000-210,000} \\P(X\geq 225,000) =0.4= 40\%[/tex]

(b)The probability he will get less than $217,000 is:

[tex]P(X\leq 217,000) =\frac{217,000-a}{b-a} =\frac{217,000-210,000}{235,000-210,000} \\P(X\leq 217,000) =0.28= 28\%[/tex]

(c) The expected value (E) and the standard deviation (S) are:

[tex]E=\frac{a+b}{2}=\frac{210,000+235,000}{2}\\ E=\$222,500\\S=\frac{b-a}{\sqrt{12}}=\frac{235,000-210,000}{\sqrt{12}}\\S=\$7,216.88[/tex]

The probabilities of the house selling for at least $225,000 and less than $217,000 are 40% and 28% respectively. The expected selling price if left on the market is $222,500 and the standard deviation is around $7,211.1.

This question is about the calculation of probabilities and expected values related to the selling price of a house. Let's solve this step by step:

We want to calculate the probability that the price of the house will be at least $225,000. The price range is uniformly distributed between $210,000 and $235,000. Therefore, to find the probability of the price being at least $225,000, we subtract the lower limit of this range ($225,000) from the upper limit ($235,000), and divide that by the entire possible price range ($235,000 - $210,000). That is $(235,000-225,000)/(235,000-210,000) = 0.4$ or 40%.

Next, to calculate the probability that the price of the house will be less than $217,000, we subtract the lower limit of the price range ($210,000) from our desired limit ($217,000), and divide that by the full price range ($235,000 - $210,000). That is $(217,000-210,000)/(235,000-210,000) = 0.28$ or 28%.

For part (c), the expected value of a uniform distribution is the midpoint of the range, or $(210,000 + 235,000)/2 = $222,500$. The standard deviation of a uniform distribution is the square root of ((upper limit - lower limit)^2/12), or sqrt[(235,000 - 210,000)^2/12] = $7,211.1$ approximately.

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