The Bruin Stock Fund sells Class A shares that have a front-end load of 4.8 percent, a 12b-1 fee of 0.42 percent, and other fees of 1.3 percent. There are also Class B shares with a 5 percent CDSC that declines 1 percent per year, a 12b-1 fee of 1.95 percent, and other fees of 1.3 percent. Assume the portfolio return is 11 percent per year. What is the value of $1 invested in each share class if your investment horizon is 3 years? Class A $ Class B $ What if your investment horizon is 20 years?

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.

Answer:

(A) Assume m is an odd integer.

Therefore m is of the m=2n-1 type where n is any number.

We have m + 6 = 2n-1 + 6 = 2n+5=2n+4 + 1 = 2(n+2) + 1, in which n + 2 is an integer, adding 6 to both ends.

Because m+6 is of the 2x + 1 type, where x =  n + 2; then m + 6 is an odd integer too.

(B) Provided that mn is an integer also. For some integer y and m, n are integers, this means mn is of the form mn = 2y.

And y = mn/2.

Therefore either m is divided by 2 or n is divided by 2 since y, m, n are all integers. To put it another way, either m is a multiple of 2 or n is a multiple of 2, which means that m is even or n is even.

Final answer:

The proposition regarding odd integers is proven by correcting the initial flawed proof, demonstrating that adding 6 to an odd integer results in another odd integer by using the correct form of an odd integer in the calculation.

Explanation:

The given proposition states that adding 6 to an odd integer results in another odd integer. The proof starts with the assumption that there exists an integer n such that m + 6 = 2n + 1.

This equation is supposed to define m + 6 as an odd integer. By isolating m, the proof continues to show that m itself is expressed in the form of 2(n - 3) + 1, indicating that m is also an odd integer.

To correct the proof, we should start with an odd integer m such that m = 2k + 1, where k is an integer. Adding 6 to m gives m + 6 = 2k + 7, which can be written as 2(k + 3) + 1, clearly showing that m + 6 is an odd integer.

Therefore, this proves the original proposition correctly.

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:

Step-by-step explanation:

Scenarios are: Sample Distribution, sampling distribution, Population distribution

sample distribution: A sample distribution contains the data of individuals of sample collected from a population.

Sampling distribution: distribution of multiple sample statistics

Population distribution: Statistics of entire population without drawing sny sample

In scenario 1, only one saple is taken from a population and that sample is plotted

In Scenario 2, multiple samples are taken from a population and means of each sample is plotted

In Scenario 3, scores of entire population is considered.

Answer:

A) Kw (37°C) = 2.12x10⁻¹⁴  

B) pH (37°C) = 6.84

Step-by-step explanation:

The following table shows the different values of Kw in the function of temperature:  

T(°C)          Kw  

0           0.114 x 10⁻¹⁴

10          0.293 x 10⁻¹⁴

20         0.681 x 10⁻¹⁴

25         1.008 x 10⁻¹⁴

30         1.471 x 10⁻¹⁴

40         2.916 x 10⁻¹⁴

50         5.476 x 10⁻¹⁴

100       51.3 x 10⁻¹⁴

A) The plot of the values above gives a straight line with the following equation:

y = -6218.6x - 11.426     (1)

where y = ln(Kw) and x = 1/T

Hence, from equation (1) we can find Kw at 37°C:

[tex] ln(K_{w}) = -6218.6 \cdot (1/(37 + 273)) - 11.426 = -31.49 [/tex]

[tex] K_{w} = e^{-31.49} = 2.12 \cdot 10^{-14} [/tex]  

Therefore, Kw at 37°C is 2.12x10⁻¹⁴  

B) The pH of a neutral solution  is:  

[tex] pH = -log([H^{+}]) [/tex]              (2)

The  hydrogen ion concentration can be calculated using the following equation:  

[tex] K_{w} = [H^{+}][OH^{-}] [/tex]        (3)

Since in pure water, the hydrogen ion concentration must be equal to the hydroxide ion concentration, we can replace [OH⁻] by [H⁺] in equation (3):

[tex] K_{w} = ([H^{+}])^{2} [/tex]

which gives:

[tex] [H^{+}] = \sqrt {K_{w}} [/tex]

Having that Kw = 2.12x10⁻¹⁴ at 37 °C (310 K), the pH of a neutral solution at this temperature is:

[tex] pH = -log ([H^{+}]) = -log(\sqrt {K_{w}}) = -log(\sqrt {2.12 \cdot 10^{-14}}) = 6.84 [/tex]

Therefore, the pH of a neutral solution at 37°C is 6.84.  

I hope it helps you!        

a. Endothermic (Kw increases with temperature).

b. pH = 7 for pure water at 50.8°C.

c. Plot ln(Kw) vs. 1/T to estimate Kw at 37.8°C.

d. pH = 7 for neutral solution at 37.8°C.

Here's the step-by-step solution:

a. Since Kw increases with temperature, indicating more ionization, the process is endothermic.

b. At 50.8°C, Kw ≈ 5.47 × 10^(-14). Since pure water is neutral, pH = 7.

c. To estimate Kw at 37.8°C:

  - Plot ln(Kw) against 1/T (in Kelvin).

  - Find the y-intercept of the plot. This intercept corresponds to ln(Kw) at 1/T = 0, which is at the temperature where T = 1/(37.8 + 273.15).

  - Take the exponential of this value to find Kw.

d. At 37.8°C, Kw ≈ 1.3 × 10^(-14). Since pure water is neutral, pH = 7.

The correct question is:

Values of Kw as a function of temperature are as follows: Temp (8C) Kw 0 1.14 3 10215 25 1.00 3 10214 35 2.09 3 10214 40. 2.92 3 10214 50. 5.47 3 10214

a. Is the autoionization of water exothermic or endothermic?

b. What is the pH of pure water at 50.8C?

c. From a plot of ln(Kw) versus 1/T (using the Kelvin scale), estimate Kw at 378C, normal physiological temperature. d. What is the pH of a neutral solution at 378C?

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 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]

See more about parameterization at brainly.com/question/14770282

Number of possible ways to arrange letters of word "possession"  = 75600.

Step-by-step explanation:

A classic counting problem is to determine the number of different ways that the letters of "possession" can be arranged.  We have, word "POSSESSION" which have 10 letters! as P,O,S,S,E,S,S,I,O,N . We have to find the number of ways letters of word "possession" can be arranged , in order to do that we will use simple logic as:

At first, we have a count of 10 letters and 10 places vacant to fill letters.

For first place we have a choice of 10 letters , after putting some letter from all 10 letters in first place now we are left with 9 places and 9 letters having 9 choices , similarly we'll be having 8 places , 8 letters and 8 choices and so on...... Therefore, Number of possible ways to arrange letters of word "possession"  = [tex]10.9.8.7.6.5.4.3.2.1[/tex] = [tex]10![/tex] ( 10 factorial ) , but there are 4 letters "s" are repeated and 2 letters "o" are repeated so we need to eliminate the similar combinations ∴ Number of possible ways = [tex]\frac{10!}{4!(2!)}[/tex] = 75600.

∴Number of possible ways to arrange letters of word "possession" = 75600

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:

[tex] P(X <9) [/tex]

And we can use the cumulative distribution function given by:

[tex] F(X) = \frac{x-a}{b-a}, a\leq X \leq b[/tex]

And using this we got:

[tex] P(X <9) = F(9) = \frac{9-3}{11-3}= 0.75[/tex]

Step-by-step explanation:

For this case we assume that X= represent the waiting times and for this case we have the following distribution:

[tex] X \sim Unif (a= 3, b =11)[/tex]

And the expected value is given by:

[tex]\mu= E(X) = \frac{a+b}{2}= \frac{3+11}{2}=7[/tex]

And the variance is given by:

[tex] Var(X) \sigma^2 = \frac{(b-a)^2}{12} = \frac{(11-3)^2}{12} = 5.333[/tex]

And we can find the deviation like this:

[tex] Sd(X) = \sqrt{5.333}= 2.309[/tex]

And we want to find this probability:

[tex] P(X <9) [/tex]

And we can use the cumulative distribution function given by:

[tex] F(X) = \frac{x-a}{b-a}, a\leq X \leq b[/tex]

And using this we got:

[tex] P(X <9) = F(9) = \frac{9-3}{11-3}= 0.75[/tex]

The probability that a randomly selected Starbucks drive-through customer in the US waits at most 9 minutes to receive their order is 0.75.

Uniform distributions are probability distributions in which all events are equally likely to occur.

Let's assume that X represents the waiting time.

As the distribution is the uniform distribution, we can write a =3 and b =11

Now, the expected value can be written as,

[tex]\mu = E(X) = \dfrac{a+b}{2} = \dfrac{3+11}{2}= 7[/tex]

the variance of the distribution can be written as,

[tex]Var(X) = \sigma^2 = \dfrac{(b-a)^2}{12} = \dfrac{(11-3)^2}{12} =5.34[/tex]

And, the standard deviation can be written as,

[tex]\sigma = \sqrt{5.34}=2.309[/tex]

In order to calculate the probability, we will use the cumulative distribution function. The cumulative function is given by the formula,

[tex]F(X) = \dfrac{x-a}{b-a}, a\leq X\leq b[/tex]

Using the function we can get the probability as,

[tex]F(X < 9) = F(9) = \dfrac{9-3}{11-3} = 0.75[/tex]

Hence,  the probability that a randomly selected Starbucks drive-through customer in the US waits at most 9 minutes to receive their order is 0.75.

Learn more about Uniform Distribution:

https://brainly.com/question/10658260

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:

There is not enough evidence to support the claim that union membership increased.  

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 400

p = 12.5% = 0.125

Alpha, α = 0.05

Number of women belonging to union , x = 52

First, we design the null and the alternate hypothesis  

[tex]H_{0}: p = 0.125\\H_A: p > 0.125[/tex]

The null hypothesis sates that 12.5% of U.S. workers belong to union and the alternate hypothesis states that there is a increase in union membership.

This is a one-tailed(right) test.  

Formula:

[tex]\hat{p} = \dfrac{x}{n} = \dfrac{52}{400} = 0.13[/tex]

[tex]z = \dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}[/tex]

Putting the values, we get,

[tex]z = \displaystyle\frac{0.13-0.125}{\sqrt{\frac{0.125(1-0.125)}{400}}} = 0.3023[/tex]

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

P-value = 0.3812

Since the p-value is greater than the significance level, we fail to reject the null hypothesis and accept the null hypothesis.

Conclusion:

Thus, there is not enough evidence to support the claim that union membership increased.

Final answer:

To determine whether union membership increased in 2006, the null hypothesis states the proportion is 12.5% or less, and the alternative hypothesis states it is greater than 12.5%. Based on the sample data (where 52 out of 400 workers are union members), the p-value is calculated to be approximately 0.22. Since the p-value is greater than 0.05, we fail to reject the null hypothesis, indicating insufficient evidence of an increase in union membership.

Explanation:

To determine whether union membership increased in 2006, we start by formulating the hypotheses for our hypothesis test:

Hypotheses

Null hypothesis (H0): The proportion of U.S. workers belonging to unions in 2006 is equal to or less than the 2005 level of 12.5% (p ≤ 0.125).

Alternative hypothesis (H1): The proportion of U.S. workers belonging to unions in 2006 is greater than the 2005 level of 12.5% (p > 0.125).

Next, we calculate the test statistic and the p-value based on the sample results:

The sample proportion is the number of workers belonging to unions divided by the total number of workers in the sample. Therefore:

Sample proportion = 52/400 = 0.13 or 13%

To find the p-value, we assume the null hypothesis is true. The test statistic for a one-sample Z-test for proportions is calculated as:

Z = (Sample proportion - Hypothesized proportion) / Standard error of the sample proportion

Standard error = sqrt((Hypothesized proportion * (1 - Hypothesized proportion)) / sample size)

Z = (0.13 - 0.125) / sqrt((0.125 * (1 - 0.125)) / 400) = 0.7727

Since this is a one-tailed test, the p-value is the probability that the standard normal variable is greater than the calculated Z value. We find the p-value using standard normal distribution tables or software. Assuming the Z value is 0.7727, the corresponding p-value would be approximately 0.22.

As the p-value is greater than typical significance levels like 0.05, we fail to reject the null hypothesis. This means there is not enough evidence at the 0.05 significance level to conclude that union membership has increased in 2006.

Answer:

145.6 seconds

Step-by-step explanation:

Mean time (μ) = 120 seconds

Standard deviation (σ) = 20 seconds

In a normal distribution, the z-score for any time, X, is given by:

 [tex]z=\frac{X-\mu}{\sigma}[/tex]

The slowest 10% correspond to the 90th percentile of a normal distribution, which has a corresponding z-score of roughly 1.28. The lowest time that requires remedial training is:

[tex]1.28=\frac{X-120}{20}\\X=145.6\ seconds[/tex]

 Times of 145.6 seconds and over qualify for remedial training.

Answer:

[tex]t=\frac{2.1-1.7}{\frac{1.01}{\sqrt{48}}}=2.744[/tex]    

[tex]p_v =P(t_{(47)}>2.744)=0.0043[/tex]  

If we compare the p value and the significance level given [tex]\alpha=0.02[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the true mean is higher than 1,7 entrees per order at 2% of signficance.  

Step-by-step explanation:

Data given and notation  

[tex]\bar X=2.1[/tex] represent the mean

[tex]s=1.01[/tex] represent the sample standard deviation

[tex]n=48[/tex] sample size  

[tex]\mu_o =1.7[/tex] represent the value that we want to test

[tex]\alpha=0.02[/tex] represent the significance level for the hypothesis 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 1.7, the system of hypothesis would be:  

Null hypothesis:[tex]\mu \leq 1.7[/tex]  

Alternative hypothesis:[tex]\mu > 1.7[/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

We can replace in formula (1) the info given like this:  

[tex]t=\frac{2.1-1.7}{\frac{1.01}{\sqrt{48}}}=2.744[/tex]    

P-value

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

[tex]df=n-1=48-1=47[/tex]  

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

[tex]p_v =P(t_{(47)}>2.744)=0.0043[/tex]  

Conclusion  

If we compare the p value and the significance level given [tex]\alpha=0.02[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the true mean is higher than 1,7 entrees per order at 2% of signficance.  

Answer:

Yes, this sample show that the average number of entrees per order was greater than expected.

Step-by-step explanation:

We are given that the  number of entrees purchased in a single order at a Noodles & Company restaurant has had an historical average of 1.7 entrees per order. For this a random sample of 48 Noodles orders had a mean number of entrees equal to 2.1 with a standard deviation equal to 1.01.

We have to test that the average number of entrees per order was greater than expected or not.

Let, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu[/tex] = 1.7 {means that the average number of entrees per order was same as expected of 1.7 entrees per order}

Alternate Hypothesis, [tex]H_1[/tex] : [tex]\mu[/tex] > 1.7 {means that the average number of entrees per order was greater than expected of 1.7 entrees per order}

The test statistics that will be used here is One sample t-test statistics;

          T.S. = [tex]\frac{Xbar-\mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]

where, Xbar = sample mean number of entrees = 2.1

              s = sample standard deviation = 1.01

              n = sample of Noodles = 48

So, test statistics = [tex]\frac{2.1-1.7}{\frac{1.01}{\sqrt{48} } }[/tex] ~ [tex]t_4_7[/tex]

                            = 2.744

Now, at 2% significance level the critical value of t at 47 degree of freedom in t table is given as 2.148. Since our test statistics is more than the critical value of t which means our test statistics will lie in the rejection region. So, we have sufficient evidence to reject our null hypothesis.

Therefore, we conclude that the average number of entrees per order was greater than expected.

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]

For more information, visit

https://brainly.com/question/14986378

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).

https://brainly.com/question/13501743

#SPJ6

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:

hope it helps you see the attachment for further information

Answer:

81

Step-by-step explanation:

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

Read more at; https://brainly.com/question/14315274

Answer:

A B D and E

Step-by-step explanation:

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:

a) [tex]Z = 1.5[/tex]

b) [tex]Z = -1.5[/tex]

c) [tex]Z = 0[/tex]

d) [tex]Z = 3[/tex]

Step-by-step explanation:

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]

In this problem, we have that:

[tex]\mu = 170, \sigma = 20[/tex]

a. 200 pounds.

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

[tex]Z = \frac{200 - 170}{20}[/tex]

[tex]Z = 1.5[/tex]

b. 140 pounds.

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

[tex]Z = \frac{140 - 170}{20}[/tex]

[tex]Z = -1.5[/tex]

c. 170 pounds.

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

[tex]Z = \frac{170 - 170}{20}[/tex]

[tex]Z = 0[/tex]

d. 230 pounds.

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

[tex]Z = \frac{230 - 170}{20}[/tex]

[tex]Z = 3[/tex]

Answer:

Yes at the level of 0.02 significance

Step-by-step explanation:

we want to compare if P₁ = P₂

P1 = 9/142= 0.0634

P2 = 5/268  = 0.0187

P = 14/410 = 0.03414

significance level, α = 0.02

Test statistic, z = [tex]\frac{p1 - p2}{\sqrt{(p * (1 - p} )) * (\frac{1}{142} * \frac{1}{268})}}[/tex]

Test Statistic, z = [tex]\frac{0.0634 - 0.0187}{\sqrt{({0.0341 * ({1 - 0.0341}})) * ({\frac{1}{142} + \frac{1}{268} ) }} } = 2.373[/tex] 

Test statistic, z = 2.373

p-value = 2*p(z<|z₀|) = 2*p(z<2.37) = 0 .0176

Answer: Since p-value (0.0176) is less than the significance level, α (0.02), the null hypothesis can not hold. we can therefore say that at 0.02 level of significance, there is sufficient evidence, statistically, that p₁ is different from p₂

Answer:

0.2843 = 28.43% probability that no less than 92 out of 159 students will pass their college placement exams.

Step-by-step explanation:

I am going to 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]p = 0.55, n = 159[/tex]. So

[tex]\mu = E(X) = 159*0.55 = 87.45[/tex]

[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{159*0.55*0.45} = 6.27[/tex]

Probability that no less than 92 out of 159 students will pass their college placement exams.

No less than 92 is more than 91, which is 1 subtracted by the pvalue of Z when X = 91. So

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

[tex]Z = \frac{91 - 87.45}{6.27}[/tex]

[tex]Z = 0.57[/tex]

[tex]Z = 0.57[/tex] has a pvalue of 0.7157

1 - 0.7157 = 0.2843

0.2843 = 28.43% probability that no less than 92 out of 159 students will pass their college placement exams.

To approximate the probability that at least 92 out of 159 students will pass their college placement exams using the normal distribution, we will follow these steps:
Step 1: Determine the parameters of the binomial distribution.
In a binomial distribution the parameters are the number of trials (n) and the probability of success in a single trial (p). For this particular case:
n = 159 (the number of students)
p = 0.55 (the probability that a given student will pass)
Step 2: Calculate the mean (μ) and standard deviation (σ) of the binomial distribution.
The mean (μ) of a binomial distribution is given by:
μ = n * p
The standard deviation (σ) is given by:
σ = sqrt(n * p * (1 - p))
For our case:
μ = 159 * 0.55 ≈ 87.45
σ = sqrt(159 * 0.55 * (1 - 0.55)) ≈ sqrt(159 * 0.55 * 0.45) ≈ sqrt(71.775) ≈ 8.472 (rounded to three decimal places for intermediate calculation)
Step 3: Apply the continuity correction.
Because we're approximating a discrete distribution with a continuous one, we use a continuity correction. To find the probability of at least 92 students passing, we'll look for the probability of X > 91.5 (since X is a discrete random variable, X ≥ 92 is equivalent to X > 91.5 when using a continuous approximation).
Step 4: Calculate the z-score for the corrected value.
The z-score is calculated by taking the value of interest, applying the continuity correction, subtracting the mean, and then dividing by the standard deviation:
z = (X - μ) / σ
Using the corrected value:
z = (91.5 - 87.45) / 8.472 ≈ 0.4774 (rounded to four decimal places for the calculation)
Step 5: Find the corresponding probability.
The z-score tells us how many standard deviations away from the mean our value of interest is. To find the probability that at least 92 students pass (i.e., P(X ≥ 92), we need to find 1 - P(Z < z) because the normal distribution table or a calculator gives us P(Z < z), which is the probability of being less than a certain z value.
We are looking for the probability that z is greater than our calculated value, which is the upper tail of the distribution.
Step 6: Consult the standard normal distribution table or use a calculator.
The z-score of approximately 0.4774 corresponds to a percentage in the standard normal distribution. Since we want P(Z > z), we need to subtract P(Z < z) from 1. If we look up the probability for z=0.4774 in standard normal distribution tables or use a calculator, we find that P(Z < z) is approximately 0.6832.
Therefore, P(Z > z) = 1 - P(Z < z) = 1 - 0.6832 = 0.3168.
Step 7: Round the answer.
Rounding P(Z > z) = 0.3168 to four decimal places, the answer remains 0.3168.
So, the approximate probability that at least 92 out of 159 students will pass the exam, using the normal distribution approximation, is 0.3168 when rounded to four decimal places.

Answer:

0.000001 = 0.0001% probability that exactly 2 of them have never been married

Step-by-step explanation:

For each employed women, there are only two possible outcomes. Either they have already been married, or they have not. The women are chosen at random, which means that the probability of a woman having been already married is independent from other women. 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.

74% of employed women have never been married.

This means that [tex]p = 0.74[/tex]

15 employed women are randomly selected

This means that [tex]n = 15[/tex]

a. What is the probability that exactly 2 of them have never been married

This is P(X = 2).

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

[tex]P(X = 2) = C_{15,2}.(0.74)^{2}.(0.26)^{13} = 0.000001[/tex]

0.000001 = 0.0001% probability that exactly 2 of them have never been married

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.

Learn more: https://brainly.com/question/1214333

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:

a)  9*x + 27

b) 9*x+3

c) 45

d) 21

Step-by-step explanation:

since (f o g)(x) = f (g(x))  , then

a) (f o g)(x) = f (x + 3) = 9*(x+3) = 9*x + 27

similarly

b) (g o f)(x) = g (f(x)) = g ( 9x) = (9*x)+3 = 9*x+3

c) for x=2

(f o g)(2) = 9*2 + 27 = 45

d) for x=2

(g o f )(2) =  9*2 +3 = 21

thus we can see that the composition of functions is not necessarily commutative  

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]

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]