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kondor19780726 [428]
3 years ago
14

What is the difference between a standard normal distribution and a nonstandard normal​ distribution? Choose the correct answer

below. A. The standard normal distribution has a standard deviation that is less than or equal to the​ mean, while a nonstandard normal distribution has a standard deviation that is greater than the mean. B. The standard normal distribution has a mean of 0 and a standard deviation of​ 1, while a nonstandard normal distribution has a different value for one or both of those parameters. C. The standard normal distribution has a standard deviation that is greater than the​ mean, while a nonstandard normal distribution has a standard deviation that is less than or equal to the mean. D. The standard normal distribution has a mean of 1 and a standard deviation of​ 0, while a nonstandard normal distribution has a different value for one or both of those parameters.
Mathematics
1 answer:
devlian [24]3 years ago
8 0

Answer:

b)The standard normal distribution has a mean of 0 and a standard deviation of​ 1, while a nonstandard normal distribution has a different value for one or both of those parameters.

Step-by-step explanation:

The standard normal distribution, ( and associated tables )  N (0,1) where 0 is the mean and 1 the standard deviation, is a model representative of all other nonstandard normal distribution. Therefore we apply such concepts and tebles in the solution of problems concerning normal distribution

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Explain how slope is used to determine if a triangle is a right triangle.
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Solve, using linear combination. <br> 3x + y = 4<br> 2x + y = 4
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----------------
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What is the solution for the system of equations? 2y=4x-12 and y=-2x+2
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A researcher wanted to see if giving a selective serotonin reuptake inhibitor (anti-depressant) would decrease the number of sel
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Answer:

a. The mean of the sample is M=35.

The variance of the sample is s^2=39.125.

The standard deviation of the sample is s=6.255.

b. z=-1.6

c. SEM = 2.212

Step-by-step explanation:

The mean of the sample is M=35.

The variance of the sample is s^2=39.125.

The standard deviation of the sample is s=6.255.

<u>Sample mean</u>

<u />M=\dfrac{1}{8}\sum_{i=1}^{8}(27+25+32+40+43+37+35+38)\\\\\\ M=\dfrac{277}{8}=35

<u>Sample variance and standard deviation</u>

<u />s^2=\dfrac{1}{(n-1)}\sum_{i=1}^{8}(x_i-M)^2\\\\\\s^2=\dfrac{1}{7}\cdot [(27-(35))^2+(25-(35))^2+(32-(35))^2+(40-(35))^2+(43-(35))^2+(37-(35))^2+(35-(35))^2+(38-(35))^2]\\\\\\

s^2=\dfrac{1}{7}\cdot [(58.141)+(92.641)+(6.891)+(28.891)+(70.141)+(5.64)+(0.14)+(11.39)]\\\\\            s^2=\dfrac{273.875}{7}=39.125\\\\\\s=\sqrt{39.125}=6.255

b. If the population mean is 45, the z-score for M=35 would be:

z=\dfrac{M-\mu}{\sigma}=\dfrac{35-45}{6.255}=\dfrac{-10}{6.255}=-1.6

c. The standard error of the mean (SEM) of this group is calculated as:

SEM=\dfrac{s}{\sqrt{n}}=\dfrac{6.255}{\sqrt{8}}=\dfrac{6.255}{2.828}=2.212

4 0
3 years ago
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