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Vinil7 [7]
3 years ago
14

Select True or False from each pull-down menu, depending on whether the corresponding statement is true or false.

Mathematics
1 answer:
Nina [5.8K]3 years ago
5 0

Answer:

1) FALSE

2) FALSE

3) TRUE

4) TRUE

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Part 1

The mean and standard deviation of a normally distributed random variable which has been standardized are one and zero, respectively.

The mean and the deviation for  the normal standard distribution are:

\mu=0, \sigma =1

So then that's FALSE.

Part 2

Using the standard normal curve, the z−score representing the 10th percentile is 1.28.

We are looking for a value a, that satisfy this:

P(X>a)=0.90   (a)

P(X   (b)  

We can find a z value that satisfy the condition with 0.10 of the area on the left and 0.90 of the area on the right it's z=-1.28. On this case P(Z<-1.28)=0.10 and P(Z>-1.28)=0.90

So that's FALSE.

Part 3

Let z1 be a z−score that is unknown but identifiable by position and area. If the symmetrical area between −z1 and +z1 is 0.9544, the value of z1 is 2.0

If we find this probability:

P(-2.0< Z

So on this case we can approximate to 0.9544. And that's TRUE.

Part 4

Using the standard normal curve, the z−score representing the 75th percentile is 0.67.

We are looking for a value a, that satisfy this:

P(X>a)=0.25   (a)

P(X   (b)  

We can find a z value that satisfy the condition with 0.75 of the area on the left and 0.25 of the area on the right it's z=0.67. On this case P(Z<0.67)=0.75 and P(Z>0.67)=0.25

So that's TRUE.

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Answer:

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Step-by-step explanation:

Here, x represents the number of 3 lb weights and y represents the number of 10 lb weights,

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