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faltersainse [42]
4 years ago
9

Which of the following functions computes a value such that 2.5% of the area under the standard normal distribution lies in the

upper tail defined by this value? ○ 0 ° ○ a.-NORMSINV(0.975) b.-NORM·S.INV(0.05) C.-NORM.S-INV(0.95) d.-NORMSINV(0.025)

Mathematics
1 answer:
fiasKO [112]4 years ago
6 0

Answer:

a) NORM.S.INV(0.975)

Step-by-step explanation:

1) Some definitions

The standard normal distribution is a particular case of the normal distribution. The parameters for this distribution are: the mean is zero and the standard deviation of one. The random variable for this distribution is called Z score or Z value.

NORM.S.INV Excel function "is used to find out or to calculate the inverse normal cumulative distribution for a given probability value"

The function returns the inverse of the standard normal cumulative distribution(a z value). Since uses the normal standard distribution by default the mean is zero and the standard deviation is one.

2) Solution for the problem

Based on this definition and analyzing the question :"Which of the following functions computes a value such that 2.5% of the area under the standard normal distribution lies in the upper tail defined by this value?".

We are looking for a Z value that accumulates 0.975 or 0.975% of the area on the left and by properties since the total area below the curve of any probability distribution is 1, then the area to the right of this value would be 0.025 or 2.5%.

So for this case the correct function to use is: NORM.S.INV(0.975)

And the result after use this function is 1.96. And we can check the answer if we look the picture attached.

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Answer:

4400

Step-by-step explanation:

15 percent is 660

this means 1 percent is 44

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3 years ago
Directions in picture.
densk [106]
Use PEMDAS, the answer that I got is 19.
8 0
3 years ago
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What is the answer to 0.02 is 1/10 of?
oee [108]

Answer:

0.2

Step-by-step explanation:

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7 0
3 years ago
Will Mark Brainlest helpppp​
Anvisha [2.4K]

Answer:

R = \left[\begin{array}{ccc}-3&-2\\1&-3\\\end{array}\right]

Step-by-step explanation:

P - Q + R = I ( I is the identity matrix )

\left[\begin{array}{ccc}2&4\\3&5\\\end{array}\right] - \left[\begin{array}{ccc}-2&2\\4&1\\\end{array}\right] + R = \left[\begin{array}{ccc}1&0\\0&1\\\end{array}\right] ( subtract corresponding elements )

\left[\begin{array}{ccc}2-(-2)&4-2\\3-4&5-1\\\end{array}\right] + R = \left[\begin{array}{ccc}1&0\\0&1\\\end{array}\right]

\left[\begin{array}{ccc}4&2\\-1&4\\\end{array}\right] + R = \left[\begin{array}{ccc}1&0\\0&1\\\end{array}\right]

R = \left[\begin{array}{ccc}1&0\\0&1\\\end{array}\right] - \left[\begin{array}{ccc}4&2\\-1&4\\\end{array}\right] = \left[\begin{array}{ccc}-3&-2\\1&-3\\\end{array}\right]

6 0
3 years ago
Find the area of the figure.
konstantin123 [22]

Answer:

<em>132</em>

<em />

Explanation:

All you need to do is separately find the area of the shapes that are part of the figure (in this case the trapezoid and rectangle), and then add them all up.

1.(7x4) + (1/2x8[7+19])

2. 28 + 104

3. 132

~Hope this helps you~

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3 years ago
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