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Hatshy [7]
3 years ago
15

Which of these variables is least likely to have a normal distribution?

Mathematics
2 answers:
Ainat [17]3 years ago
8 0

Answer:

Annual income for all 150 employees at a local high school

Step-by-step explanation:

The normal distribution is also called the Gaussian distribution. It is the most common type of distribution for statistical analysis. The normal distribution is based on a probability distribution. The data near the mean are more frequent than the data away from the mean. It is based on two parameters: the mean and the standard deviation.

As per options, option A which states ‘annual income for all 150 employees at a local high school’ the data is within the standard deviation.  

Viefleur [7K]3 years ago
4 0
It would be a, because some of the students could have really high income and some low, it’s very unpredictable, but all others will probably all be very close

Hope this helps!
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Plz help me on this
Margarita [4]
$125.4 is the answer to #1
#2 im not sure how the 10% effects the problem. so i cant help you there
7 0
3 years ago
5. Find the quotient: 0.382 +0.2.<br> A 0.0764<br> C 1.91<br> B 0.764<br> D 0.191
ankoles [38]

Answer:

The expression is not correctly given. However If you want the quotient of 0.2 divided by 0.382. This would be the correct answer

C.  \: 1.91

Hope thai helps!

please like if I'd does help! (:

3 0
3 years ago
Which of the following points lies on the graph of the function y= 2.3*?
Rainbow [258]

If the equation is y = 2\cdot 3^x  (2 times 3^x)

Then the answer is d. (0,2)

This is because plugging x = 0 leads to y = 2 as shown below

y = 2\cdot 3^x\\\\y = 2\cdot 3^0\\\\y = 2\cdot 1\\\\y = 2

3 0
3 years ago
A random sample of 49 lunch customers was taken at a restaurant. The average amount of time the customers in the sample stayed i
Hunter-Best [27]

Answer:

a)  σ/√n= 1.43 min

c) Margin of error 2.8028min

d) [30.1972; 35.8028]min

e) n=62 customers

Step-by-step explanation:

Hello!

The variable of interest is

X: Time a customer stays at a restaurant. (min)

A sample of 49 lunch customers was taken at a restaurant obtaining

X[bar]= 33 mi

The population standard deviation is known to be δ= 10min

a) and b)

There is no information about the distribution of the population, but we know that if the sample is large enough, n≥30, we can apply the central limit theorem and approximate the distribution of the sample mean to normal:

X[bar]≈N(μ;σ²/n)

Where μ is the population mean and σ²/n is the population variance of the sampling distribution.

The standard deviation of the mean is the square root of its variance:

√(σ²/n)= σ/√n= 10/√49= 10/7= 1.428≅ 1.43min

c)

The CI for the population mean has the general structure "Point estimator" ± "Margin of error"

Considering that we approximated the sampling distribution to normal and the standard deviation is known, the statistic to use to estimate the population mean is Z= (X[bar]-μ)/(σ/√n)≈N(0;1)

The formula for the interval is:

[X[bar]±Z_{1-\alpha /2}*(σ/√n)]

The margin of error of the 95% interval is:

Z_{1-\alpha /2}= Z_{1-0.025}= Z_{0.975}= 1.96

d= Z_{1-\alpha /2}*(σ/√n)= 1.96* 1.43= 2.8028

d)

[X[bar]±Z_{1-\alpha /2}*(σ/√n)]

[33±2.8028]

[30.1972; 35.8028]min

Using a confidence level of 95% you'd expect that the interval [30.1972; 35.8028]min contains the true average of time the customers spend at the restaurant.

e)

Considering the margin of error d=2.5min and the confidence level 95% you have to calculate the corresponding sample size to estimate the population mean. To do so you have to clear the value of n from the expression:

d= Z_{1-\alpha /2}*(σ/√n)

\frac{d}{Z_{1-\alpha /2}}= σ/√n

√n*(\frac{d}{Z_{1-\alpha /2}})= σ

√n= σ* (\frac{Z_{1-\alpha /2}}{d})

n=( σ* (\frac{Z_{1-\alpha /2}}{d}))²

n= (10*\frac{1.96}{2.5})²= 61.47≅ 62 customers

I hope this helps!

3 0
3 years ago
Hi!! So I am doing test corrections for a test that I did bad on but I need to know what is the equation for this??
Sedaia [141]

Answer:

u can use fraction calculator

Step-by-step explanation:

just go and search 'fraction calculator'

8 0
2 years ago
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