Find the area of the largest rectangle that can be inscribed in a right triangle with legs of lengths 4 cm and 6 cm if two sides of the rectangle lie along the legs. webassign cengage

Answer:

41 word/min

Step-by-step explanation:

Before noon Ali works:

She types:

After lunch she works:

She types:

Total Ali works= 4+4= 8 hours= 480 min

Total Ali types= 11520+8160= 19680 words

Average typing rate= 19680 words/480 min= 41 word/min

Answer: A. This is a discrete probability distribution.

hours             probability

  1                      0.09        

 2                      0.16

 3                      0.23

 4                      0.17

 5                      0.09

 6                      0.05

 7                      0.04

 8                      0.16

B. E(X) = 4.12; σ = 2.21

C. μ = 12.75; s = 6.11

Step-by-step explanation: Probability Distribution is an equation or table linking each outcome of an experiment with its probability of ocurrence. For this case, since the experiment is performed a high number of times and in a long run, the relative frequency of the event is its probability. Therefore:

A. To convert to a probability distribution, find the probability through the frequency by doing:

Hour 1

P(X) = [tex]\frac{21}{228}[/tex] = 0.09

Hour 2

P(X) = [tex]\frac{36}{228}[/tex] = 0.16

Hour 3

P(X) = [tex]\frac{53}{228}[/tex] = 0.23

Hour 4

P(X) = [tex]\frac{40}{228}[/tex] = 0.17

Hour 5

P(X) = [tex]\frac{22}{228}[/tex] = 0.09

Hour 6

P(X) = [tex]\frac{11}{228}[/tex] = 0.05

Hour 7

P(X) = [tex]\frac{9}{228}[/tex] = 0.04

Hour 8

P(X) = [tex]\frac{36}{228}[/tex] = 0.16

The table will be:  

hours             probability

  1                      0.09        

 2                      0.16

 3                      0.23

 4                      0.17

 5                      0.09

 6                      0.05

 7                      0.04

 8                      0.16

This is a discrete distribution because it lists all the possible values that the discrete variable can be and its associated probabilities.

B. Mean for a probability distribution is calculated as:

E(X) = ∑[[tex]x_{i}[/tex].P([tex]x_{i}[/tex])]

E(X) = 1*0.09 + 2*0.16+3*0.23+4*0.17+5*0.09+6*0.05+7*0.04+8*0.16

E(X) = 4.12

Standard Deviation is:

σ = √∑{[x - E(x)]² . P(x)}

σ = [tex]\sqrt{(1-4.12)^{2}*0.09 + (2-4.12)^{2}*0.16 + ... + (7-4.12)^{2}*0.04 + (8-4.12)^{2}*0.16}[/tex]

σ = [tex]\sqrt{4.87}[/tex]

σ = 2.21

The average number of hours parked is approximately 4h with a standard deviation of approximately 2 hours, which means that a typical costumer parks between 2 to 6 hours.

C. Mean for a sample is given by: μ = ∑[tex]\frac{x_{i}}{n}[/tex] , which is this case is:

μ = [tex]\frac{4+6+9+13+14+16+18+22}{8}[/tex]

μ = 12.75

Standard Deviation of a sample: s = √[tex]\frac{1}{n-1}[/tex]∑([tex]x_{i}[/tex] - μ)²

s =  [tex]\sqrt{ \frac{(4-12.75)^{2} + (6-12.74)^{2} + ... + (18-12.75)^{2} + (22-12.75)^{2} }{8-1}}[/tex]

s = 6.11

The average amount charged is 12.75±6.11.

Answer: (fоg)(24)=5

Step-by-step explanation:

(fоg)(24) is f of g of 24. This means you plug in g(24) into f(x).

[tex]g(24)=\sqrt{24-8}[/tex]

[tex]g(24)=\sqrt{16}[/tex]

[tex]g(24)=4[/tex]

Now that we know g(24), we can plug it into f(x).

f(4)=2(4)-3

f(4)=8-3

f(4)=5

Answer:

5/33649= approx 0.00015

Step-by-step explanation:

Total number of outcomes are  C24 6= 24!/(24-6)!/6!=19*20*21*22*23*24/(2*3*4*5*6)= 19*14*22*23

Half of the Committee =3 persons. That mens that number of the women in Commettee=3.  3 women from 6 can be elected C6  3  ways ( outputs)=

6!/3!/3!=4*5*6*/2/3=20

So the probability that 3 members of the commettee are women  is

P(women=3)= 20/(19*14*22*23)=5/(77*19*23)=5/33649=approx 0.00015

The probability that precisely half of the members will be women is;

P(3 women) = 0.1213

This question will be solved by hypergeometric distribution which has the formula;

P(x) = [S_C_s × (N - S)_C_(n - s)]/(NC_n)

where;

S is success from population

s is success from sample

N is population size

n is sample size

We are give;

s = 3 women (which is precisely half of the members selected)

S = 6 women

N = 24 men and women

n = 6 people selected

Thus;

P(3 women) = (⁶C₃ * ⁽¹⁸⁾C₍₃₎)/(²⁴C₆)

P(3 women) = (20 * 816)/134596

P(3 women) = 0.1213

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Answer:

The third options

Step-by-step explanation:

Counting we can see that 10 students went to two or less states, and 10 went to three or more

Answer:

50,063,860 different samples can be chosen

Step-by-step explanation:

The order in which the microchips are chosen is not important. So we use the combinations formula to solve this question.

Combinations formula:

[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]

How many different samples can be chosen

We choose 6 microchips from a set of 60. So

[tex]C_{60,6} = \frac{60!}{6!(60-6)!} = 50063860[/tex]

50,063,860 different samples can be chosen

Answer:

Step-by-step explanation:

From the information given, the population is divided into sub groups. Each group would consist of citizens picking a particular choice as the most important problem facing the country. The choices are the different categories. In this case, the null hypothesis would state that the distribution of proportions for all categories is the same in each population. The alternative hypothesis would state that the distributions is different. Therefore, the correct test to use to determine if the distribution of​ "problem facing this country ​today" is different between the two different​ years is

A) Use a​ chi-square test of homogeneity.

Answer:

1)

A) [tex]\frac{}{X}[/tex] ~ N(8.5;0.108)

B) P([tex]\frac{}{X}[/tex] > 9)= 0.0552

C) P(X> 9)= 0.36317

D) IQR= 0.4422

2)

A) [tex]\frac{}{X}[/tex] ~ N(30;2.5)

B) P( [tex]\frac{}{X}[/tex]<30)= 0.50

C) P₉₅= 32.60

D) P( [tex]\frac{}{X}[/tex]>36)= 0

E) Q₃: 31.0586

Step-by-step explanation:

Hello!

1)

The variable of interest is

X: pollutants found in waterways near a large city. (ppm)

This variable has a normal distribution:

X~N(μ;σ²)

μ= 8.5 ppm

σ= 1.4 ppm

A sample of 18 large cities were studied.

A) The sample mean is also a random variable and it has the same distribution as the population of origin with exception that it's variance is affected by the sample size:

[tex]\frac{}{X}[/tex] ~ N(μ;σ²/n)

The population mean is the same as the mean of the variable

μ= 8.5 ppm

The standard deviation is

σ/√n= 1.4/√18= 0.329= 0.33 ⇒σ²/n= 0.33²= 0.108

So: [tex]\frac{}{X}[/tex] ~ N(8.5;0.108)

B)

P([tex]\frac{}{X}[/tex] > 9)= 1 - P([tex]\frac{}{X}[/tex] ≤ 9)

To calculate this probability you have to standardize the value of the sample mean and then use the Z-tables to reach the corresponding value of probability.

Z= [tex]\frac{\frac{}{X} - Mu}{\frac{Sigma}{\sqrt{n} } } = \frac{9-8.5}{0.33}= 1.51[/tex]

Then using the Z table you'll find the probability of

P(Z≤1.51)= 0.93448

Then

1 - P([tex]\frac{}{X}[/tex] ≤ 9)= 1 - P(Z≤1.51)= 1 - 0.93448= 0.0552

C)

In this item, since only one city is chosen at random, instead of working with the distribution of the sample mean, you have to work with the distribution of the variable X:

P(X> 9)= 1 - P(X ≤ 9)

Z= (X-μ)/δ= (9-8.5)/1.44

Z= 0.347= 0.35

P(Z≤0.35)= 0.63683

Then

P(X> 9)= 1 - P(X ≤ 9)= 1 - P(Z≤0.35)= 1 - 0.63683= 0.36317

D)

The first quartile is the value of the distribution that separates the bottom 2% of the distribution from the top 75%, in this case it will be the value of the sample average that marks the bottom 25% symbolically:

Q₁: P([tex]\frac{}{X}[/tex]≤[tex]\frac{}{X}[/tex]₁)= 0.25

Which is equivalent to the first quartile of the standard normal distribution. So first you have to identify the first quartile for the Z dist:

P(Z≤z₁)= 0.25

Using the table you have to identify the value of Z that accumulates 0.25 of probability:

z₁= -0.67

Now you have to translate the value of Z to a value of [tex]\frac{}{X}[/tex]:

z₁= ([tex]\frac{}{X}[/tex]₁-μ)/(σ/√n)

z₁*(σ/√n)= ([tex]\frac{}{X}[/tex]₁-μ)

[tex]\frac{}{X}[/tex]₁= z₁*(σ/√n)+μ

[tex]\frac{}{X}[/tex]₁= (-0.67*0.33)+8.5=  8.2789 ppm

The third quartile is the value that separates the bottom 75% of the distribution from the top 25%. For this distribution, it will be that value of the sample mean that accumulates 75%:

Q₃: P([tex]\frac{}{X}[/tex]≤[tex]\frac{}{X}[/tex]₃)= 0.75

⇒ P(Z≤z₃)= 0.75

Using the table you have to identify the value of Z that accumulates 0.75 of probability:

z₃= 0.67

Now you have to translate the value of Z to a value of [tex]\frac{}{X}[/tex]:

z₃= ([tex]\frac{}{X}[/tex]₃-μ)/(σ/√n)

z₃*(σ/√n)= ([tex]\frac{}{X}[/tex]₃-μ)

[tex]\frac{}{X}[/tex]₃= z₃*(σ/√n)+μ

[tex]\frac{}{X}[/tex]₃= (0.67*0.33)+8.5=  8.7211 ppm

IQR= Q₃-Q₁= 8.7211-8.2789= 0.4422

2)

A)

X ~ N(30,10)

For n=4

[tex]\frac{}{X}[/tex] ~ N(μ;σ²/n)

Population mean μ= 30

Population variance σ²/n= 10/4= 2.5

Population standard deviation σ/√n= √2.5= 1.58

[tex]\frac{}{X}[/tex] ~ N(30;2.5)

B)

P( [tex]\frac{}{X}[/tex]<30)

First you have to standardize the value and then look for the probability:

Z=  ([tex]\frac{}{X}[/tex]-μ)/(σ/√n)= (30-30)/1.58= 0

P(Z<0)= 0.50

Then

P( [tex]\frac{}{X}[/tex]<30)= 0.50

Which is no surprise since 30 y the value of the mean of the distribution.

C)

P( [tex]\frac{}{X}[/tex]≤ [tex]\frac{}{X}[/tex]₀)= 0.95

P( Z≤ z₀)= 0.95

z₀= 1.645

Now you have to reverse the standardization:

z₀= ([tex]\frac{}{X}[/tex]₀-μ)/(σ/√n)

z₀*(σ/√n)= ([tex]\frac{}{X}[/tex]₀-μ)

[tex]\frac{}{X}[/tex]₀= z₀*(σ/√n)+μ

[tex]\frac{}{X}[/tex]₀= (1.645*1.58)+30= 32.60

P₉₅= 32.60

D)

P( [tex]\frac{}{X}[/tex]>36)= 1 - P( [tex]\frac{}{X}[/tex]≤36)= 1 - P(Z≤(36-30)/1.58)= 1 - P(Z≤3.79)= 1 - 1 = 0

E)

Q₃: P([tex]\frac{}{X}[/tex]≤[tex]\frac{}{X}[/tex]₃)= 0.75

⇒ P(Z≤z₃)= 0.75

z₃= 0.67

z₃= ([tex]\frac{}{X}[/tex]₃-μ)/(σ/√n)

z₃*(σ/√n)= ([tex]\frac{}{X}[/tex]₃-μ)

[tex]\frac{}{X}[/tex]₃= z₃*(σ/√n)+μ

[tex]\frac{}{X}[/tex]₃= (0.67*1.58)+30= 31.0586

Q₃: 31.0586

Answer:

9.42 in²

Step-by-step explanation:

The area of whole circle S=pi*R²    , where pi is appr. 3.14,  R= 6 in

S= 3.14*6² =113.04 in²

The area of smaller sector is Ssec=S/360*30=113,04/12=9.42 in²

The area of the smaller sector with a central angle of 30 degrees and a radius of 6 inches is 9.42478 square inches.

To find the area of a sector, you can use the formula:

Area of sector = (θ/360) × π × r²

where θ is the central angle in degrees, r is the radius of the sector.

The central angle is 30 degrees and the radius is 6 inches.

Plugging these values into the formula:

Area of sector = (30/360) × π × 6²

= (1/12) × π × 36

= (1/12) × 3.14159 × 36

= 9.42478 square inches

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Answer:

Step-by-step explanation:

a. The hypotheses are:

Null hypothesis: the average test scores are the same for the different teaching methods.

Alternative hypothesis: the average test scores are different for the different teaching methods.

b. To determine the degree of freedom for the F test: we must find two sources of variation such that we have two variances. The two sources of variation are: Factor (between groups) and the error (within groups) and add this up. Or use (N - 1). N is number in sample

c. With a p value of of 0.0168 and using a standard significance level of 0.05, we will reject the null hypothesis as 0.0168 is less than 0.05 and conclude that the average test scores are different for the different teaching methods.

Answer:

Option C is correct.

The sampling distribution with sample size n=100 will have less variability.

Step-by-step explanation:

Complete Question

Consider random samples selected from the population of all female college soccer players in the United States. Assume the mean height of female college soccer players in the United States is 66 inches and the standard deviation is 3.5 inches. Which do you expect to have less variability (spread): a sampling distribution with sample size n = 100 or a sample size of n = 20.

A. Both sampling distributions will have the same variability.

B.The sampling distribution with sample size n=20 will have less variability

C. The sampling distribution with sample size n =100 will have less variability

Solution

The central limit theorem allows us to say that as long as

- the sample is randomly selected from the population distribution with each variable independent of each other and with the sample having an adequate enough sample size.

- the random sample is normal or almost normal which is guaranteed if the population distribution that the random sample was extracted from is normal or approximately normal,

1) The mean of sampling distribution (μₓ) is approximately equal to the population mean (μ)

μₓ = μ = 66 inches

2) The standard deviation of the sampling distribution or the standard error of the sample mean is related to the population standard deviation through

σₓ = (σ/√N)

where σ = population standard deviation = 3.5 inches

N = Sample size

And the measure of variability for a sampling distribution is the magnitude of the standard deviation of the sampling distribution.

For sampling distribution with sample size n = 100

σₓ = (3.5/√100) = 0.35 inch

For sampling distribution with sample size n = 20

σₓ = (3.5/√20) = 0.7826 inch

The standard deviation of the sampling distribution with sample size n = 20 is more than double that of the sampling distribution with sample size n = 100, hence, it is evident that the bigger the sample size, the lesser the standard deviation of the sampling distribution and the lesser the variability that the sampling distribution shows.

Hope this Helps!!!

Answer:

yes

Step-by-step explanation:

not every person is going to have the same opinion, so it is yes.

// have a great day //

Answer:

Yes, because if Pedro asked them the question "what do you think of public transportation?" the majority would probably say that they like it or something along those lines. This is biased because there may be other city inhabitants who don't think very highly of public transportation. Basically, what I'm trying to say is that not everyone will have the same opinion.

Answer: 1/8

Step-by-step explanation:

Unit Fractions: A unit fraction is a rational number written as a fraction where the numerator is one and the denominator is a positive integer. A unit fraction is therefore the reciprocal of a positive integer, 1/n.

Example of Unit Fractions: 1/1, 1/2, 1/3, 1/4 ,1/5, etc.

Hope this helps! Please mark as brainliest!

The unit fraction of the cake is 1/8

A unit fraction is a rational number written as a fraction where the numerator is one and the denominator is a positive integer.

A unit fraction is therefore the reciprocal of a positive integer, 1/n.

Examples are 1/1, 1/2, 1/3, 1/4, 1/5, etc.

Given that, George cut a cake into 8 equal pieces, we need to find the unit fraction for the cake

Since, George cut the cake in 8 equal pieces so, 1 part will be shown by 1/8 of the cake, that mean 1/8 is one unit of the cake, we can say that 1/8 is the unit of the whole cake.

Hence, the unit fraction of the cake is 1/8

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Answer:

b: a over b divided by do over c

Step-by-step explanation:

You can solve this by plugging in numbers for each variable.

For example: a=1, b=4, c=1, d=2

1/4 ÷ 1/2 = 0.125

If you plug in the numbers for all the equations listed, only 1/4 ÷ 2/1 = 0.125.

Answer:

1/2 (simplified)

Step-by-step explanation:

6 numbers (that's the total probability) --> 6 denominator

3 are odd (odd numbers in the probability) --> 3 numerator

so => 3/6

--> simplify

1/2

Hope this helps!

Answer:

90

Step-by-step explanation:

1/1111= 0. (0009) cycles of 0009 after decimal point (one 9 per 4 digits)

Number of digits 9:

Answer:

90

Step-by-step explanation:

Answer:

a) Null hypothesis (H0): the mean guest bill for a weekend is $600.

Alternative hypothesis (Ha): the mean guest bill for a weekend is significantly bigger than $600.

b) When H0 can not be rejected, the conclusion is that there is no enough evidence to claim that the mean guest bill had increased from $600.

c) When the H0 is rejected, they have enough evidence to claim that the mean guest bill is significantly bigger than $600.  

Step-by-step explanation:

a) The accountant, as he wants to see if there is evidence to support the claim that the mean guest bill has increased significanty, should write the hypothesis like that:

Null hypothesis (H0): the mean guest bill for a weekend is $600.

Alternative hypothesis (Ha): the mean guest bill for a weekend is significantly bigger than $600.

A sample of bills of the period in study needs to be taken in order to have a representation of the actual population of bills and then perform a t-test, as the sample mean and standard deviation will be used to perform the test.

b) When H0 can not be rejected, the conclusion is that there is no enough evidence to claim that the mean guest bill had increased from $600. If the P-value was low but not enough, they may take another sample to perform the test again or leave it like that.

c) When the H0 is rejected, they have enough evidence to claim that the mean guest bill is significantly bigger than $600.  

Answer:

after the 1st year

Step-by-step explanation:

$2300 × 75% = $1725.00

$2300-$1725= $575

Answer:

9.375 in^2

Step-by-step explanation:

Answer:

[tex]\frac{1}{6}[/tex]

Step-by-step explanation:

Answer:

Step-by-step explanation:

[tex]\frac{5}{6}-\frac{2}{3}=\frac{5}{6}-\frac{2*2}{3*2}\\\\=\frac{5}{6}-\frac{4}{6}\\\\=\frac{5-4}{6}\\\\=\frac{1}{6}[/tex]

Answer:

Height of tomato plant is the dependent variable

Presence of walnut leaves in the soil is the independent variable

Tomato plants grown without walnut leaves is the control

Step-by-step explanation:

An independent variable is the variable in an experiment that can be altered to test for a certain result. It is independent, or does not change with change in other factors in the experiment. In this case, the presence or absence, or quantity of walnut available in the soil is the independent variable in the experiment.

A dependent variable varies, and depends on the independent variable. It is what is measured in the experiment. In this case, the height of the tomato plants is the dependent variable that depends on the presence, absence or quantity of walnut in the soil.

A control in an experiment, is a replicate experiment, that is manipulated in order to be able to test a single variable at a time. Controls are variables are held constant so as to minimize their effect on the system under study. In this case, some of the tomato plants are planted without walnut in the soil, to test the effect of the absence of the walnut in the soil.

Answer:

I believe it is Inductive Reasoning.

Step-by-step explanation:

Inductive Reasoning is a type of logical thinking that involves forming generalizations based on specific incidents you've experienced, observations you've made, or facts you know to be true or false.

Deductive Reasoning is a basic form of valid reasoning.

Answer:

[tex]\boxed{ \ dY_t=(2\theta+2\psi Y_t+\phi^2)dt+2\phi \sqrt{Y_t}dW_t\ }[/tex]

Step-by-step explanation:

it is a long time I have not applied Ito's lemma

I would say the following

for [tex]f(x)=x^2[/tex]

f'(x)=2x

f''(x)=2

so using Ito's lemma we can write that

[tex]dY_t=2V_tdV_t+\phi^2dt[/tex]

[tex]dY_t=2(\theta+\psi V_t^2)dt+2\phi V_tdW_t+\phi^2dt[/tex]

[tex]dY_t=(2\theta+2\psi V_t^2+\phi^2)dt+2\phi V_tdW_t[/tex]

so it comes

[tex]dY_t=(2\theta+2\psi Y_t+\phi^2)dt+2\phi \sqrt{Y_t}dW_t[/tex]

Answer:

The probability that demand during lead-time will exceed 20 bails is 0.2033.

Step-by-step explanation:

We are given that it has been previously determined that demand during the lead-time is normally distributed with a mean of 15 bails and a standard deviation of 6 bails.

Let X = demand during the lead-time

So, X ~ Normal([tex]\mu=15, \sigma^{2} = 6^{2}[/tex])

The z-score probability distribution for the normal distribution is given by;

                               Z  =  [tex]\frac{X-\mu}{\sigma}[/tex]  ~ N(0,1)

where, [tex]\mu=[/tex] population mean demand = 15 bails

           [tex]\sigma[/tex] = standard deviation = 6 bails

Now, the probability that demand during lead-time will exceed 20 bails is given by = P(X > 20 bails)

       P(X > 20 bails) = P( [tex]\frac{X-\mu}{\sigma}[/tex] > [tex]\frac{20-15}{6}[/tex] ) = P(Z > 0.83) = 1 - P(Z [tex]\leq[/tex] 0.83)

                                                             = 1 - 0.7967 = 0.2033

Answer:

The answer is B

Step-by-step explanation:

Hope this helps.. if not im sorry :(

Answer:

114

Step-by-step explanation:

Answer:

144Step-by-step explanation:

Answer:

slope = -4/5

Step-by-step explanation:

A line passes two points (x1, y1) and (x2, y2).

The slope of this line can be calculate by the formula:

s = (y2 - y1)/(x2 - x1)

=>The line that passes A(-4, 8) and B(-9, 12) has the slope:

s = (12 - 8)/(-9 - -4) = 4/(-5) = -4/5

Hope this helps!

Answer:

-30000/100

300 0/1

Step-by-step explanation:

We have the following numbers -18 and 16 2/3, the first is an integer and the second is a mixed number, the first thing is to pass the mixed number to a decimal number.

16 2/3 = 16.67

We do the multiplication:

−18⋅16 2/3 = -300

We have an improper fraction is a fraction in which the numerator (top number) is greater than or equal to the denominator (bottom number), therefore it would be:

-30000/100

How mixed number would it be:

300 0/1

The Laplace transform of the function [tex]\frac{1}{2} (\frac{1}{s} - \frac{s}{s^2 + 4w^2} )[/tex] .

The Laplace transform exist when s > 0 .

Here, the given function is f(t) = sin²(wt) .

The Laplace transform of the the function f(t),

F(s) = f(t) = { [tex]{\frac{1}{2} \times 2sin^2(wt) }[/tex] }

F(s) = { [tex]\frac{1}{2} \times (1- cos2wt)[/tex] }

F(s) = { [tex]\frac{1}{2} - \frac{1}{2} \times cos(2wt)\\[/tex] }

F(s) = [tex]\frac{1}{2} (\frac{1}{s} - \frac{s}{s^2 + 4w^2} )[/tex]

Next,

The above Laplace transform exist if s > 0 .

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Answer:

[tex] SE =\sqrt{\frac{p(1-p)}{n}}[/tex]

And replacing we got:

[tex] SE=\sqrt{\frac{0.75*(1-0.75)}{900}}= 0.014[/tex]

Step-by-step explanation:

For this case we have the following info given:

[tex] n=900[/tex] represent the sample size selected

[tex]p = 0.75[/tex] represent the population proportion

We want to find the standard error and we can use the distribution for the sample proportion and for this case since the sample size is large enough and we satisfy np>10 and n(1-p) >10 we have:

[tex] \hat p \sim N (p,\sqrt{\frac{p(1-p)}{n}})[/tex]

And the standard error is given;

[tex] SE =\sqrt{\frac{p(1-p)}{n}}[/tex]

And replacing we got:

[tex] SE= \sqrt{\frac{0.75* (1-0.75)}{900}}= 0.014[/tex]