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A line for tickets to a Broadway show had a mean waiting time of 20 minutes with a standard deviation of 5 minutes.

andrezito [222]

Answer:

5.48% of the people in line waited for more than 28 minutes

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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 p-value, we get the probability that the value of the measure is greater than X.

Mean waiting time of 20 minutes with a standard deviation of 5 minutes.

This means that $\mu = 20, \sigma = 5$

What percentage of the people in line waited for more than 28 minutes?

The proportion is 1 subtracted by the p-value of Z when X = 28. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{28 - 20}{5}$

$Z = 1.6$

$Z = 1.6$ has a p-value of 0.9452.

1 - 0.9452 = 0.0548.

As a percentage:

0.0548*100% = 5.48%

5.48% of the people in line waited for more than 28 minutes

6 0

3 years ago

Use a t-distribution to find a confidence interval for the difference in means μd=μ1-μ2 using the relevant sample results from p

V125BC [204]

Answer:

a)

best estimate = Xd[bar]=4.80

margin of error = 8.66

The 99% confidence interval is -3.86 to 13.46

b)

test statistic = -1.86

p-value = 0.0526

Decision: Reject the null hypothesis.

At the 5% significance level, you can conclude that the population mean of the difference between treatment 1 and treatment 2 is less than zero.

Step-by-step explanation:

Hello!

a) 99% CI

Using d=X₁-X₂ to determine the study variable Xd: the difference between treatment 1 and treatment 2.

Assuming that this variable has an approximately normal distribution: Xd≈N(μd;σ²d)

To calculate the sample mean and standard deviation you have to calculate the difference between the values of both treatments first.

Case 1 ; Case 2 ; Case 3 ; Case 4 ; Case 5

22-18= 4 ; 27-29= -2 ; 32-25= 7; 26-20= 6 ; 29-20= 9

n= 5

Xd[bar]= ∑X/n= 24/5= 4.80

Sd²= 1/(n-1)*[∑X²-(∑X)²/n]= 1/4*[186-(24²)/5]= 17.7

Sd= 4.21

The parameter of interes is the population mean od the difference, μd

The best estimate for this parameter is the sample mean, Xd[bar]=4.80

Using the t-distribution, the formula for the Confidence Interval is

Xd[bar] ± $t_{n-1;1-\alpha /2}$ * $\frac{Sd}{\sqrt{n} }$

Where the margin of error is:

$t_{n-1;1-\alpha /2}$ * $\frac{Sd}{\sqrt{n} }$ = $t_{4;0.995}$ * $\frac{Sd}{\sqrt{n} }$ = 4.604* $\frac{4.21}{\sqrt{5} }$ = 8.66

99% CI [-3.86; 13.46]

b) 5% Hypothesis test

The variable of interest is defined d=X₁-X₂; Xd: the difference between treatment 1 and treatment 2. Xd≈N(μd;σ²d)

The statistic hypotheses are:

H₀: μd = 0

H₁: μd < 0

α: 0.05

The statistic to use for this test is:

$t_{H_0}= \frac{X_d[bar]-Mu_d}{\frac{Sd}{\sqrt{n} } } ~~t_{n-1}$

As before you have to calculate the difference between the observation for each case and then the sample mean and standard deviation:

Case 1 ; Case 2 ; Case 3 ; Case 4 ; Case 5 ; Case 6 ; Case 7 ; Case 8

18-18= 0; 12-19= -7; 11-25= -14; 21-21= 0; 15-19= -4; 11-14=-3; 14-15= -1; 22-20= 2

n= 8

Xd[bar]= ∑X/n= -27/8= -3.38

Sd²= 1/(n-1)*[∑X²-(∑X)²/n]= 1/7*[275-(-27²)/8]= 26.27

Sd= 5.13

$t_{H_0}= \frac{-3.38-0}{\frac{5.13}{\sqrt{8} } }= -1.86$

This test is one-tailed to the left, which means that you will reject the null hypothesis to small values of t, the p-value of the test has the same direction as the rejection region, this means that it is one-tailed to the left and you can calculate it as:

P(≤-1.86)= 0.0526

The decision rule using the p-value is:

If p-value > α, do not reject the null hypothesis.

If p-value ≤ α, reject the null hypothesis.

The p-value is greater than the significance level so the decision is to reject the null hypothesis.

I hope it helps!

7 0

4 years ago

<img src="https://tex.z-dn.net/?f=f%28x%29%20%3D%20%5Cfrac%7Bx-5%7D%7B2x%5E%7B2%7D-5x-3%20%7D" id="TexFormula1" title="f(x) = \f

Vedmedyk [2.9K]

i) The given function is

$f(x)=\frac{x-5}{2x^2-5x-3}$

The factored form is

$f(x)=\frac{x-5}{(x-3)(2x+1)}$

The domain are the values of x for which the function is defined.

$(x-3)(2x+1)\ne 0$

$(x-3)\ne0,(2x+1)\ne 0$

$x\ne3,x\ne-\frac{1}{2}$

ii) To find the vertical asymptotes, equate the denominator to zero.

$(x-3)(2x+1)=0$

$(x-3)=\ne0,(2x+1)=0$

$x=3,x=-\frac{1}{2}$

iii) To find the roots, equate the numerator to zero.

$x-5=0$

The root is $x=5$

iv) To find the y-intercept, put $x=0$ into the function.

$f(0)=\frac{0-5}{(0-3)(2(0)+1)}$

$f(0)=\frac{-5}{(-3)(1)}$

$f(0)=\frac{5}{3}$

The y-intercept is $\frac{5}{3}$

v) The horizontal asymptote is given by;

$lim_{x\to \infty}\frac{x-5}{2x^2-5x-3}=0$

The horizontal asymptote is $y=0$

vi) The function is not reducible. There are no holes.

vii) The given function is a proper rational function.

Proper rational functions do not have oblique asymptotes.

3 0

3 years ago

Read 2 more answers

Which is the equation of the given line. (-4,-3) and (2,-3)

Kruka [31]

Little tip....notice how ur y coordinates are the same....this means that u have a horizontal line which is represented by y = a number...with that number being the y coordinate of ur points.
So ur equation is : y = -3 <===

** if ur x's would have been the same, it would be a vertical line represented by x = a number...with that number being the x coordinates of ur points...but thats only if ur x coordinates were the same

8 0

3 years ago

What is the surface area of a rectangular prisim with dimensions 5ft, 6ft, and 7ft.

djyliett [7]

Ok the formula for this is 2(lw +hw+hL)

2(30+35+42)=2(107)=214 square feet

8 0

3 years ago

Read 2 more answers