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Alex777 [14]
3 years ago
13

Jenny asked 10 students how many courses they have taken so far at her collegeHere is the list of answers, 12, 10, 14, 7, 1, 19,

17, 3, 20, 16 What is the percentage of these students who have taken fewer than 16 courses?

Mathematics
1 answer:
olga_2 [115]3 years ago
5 0
60%, she asked 10 students and 6 of them had less than 16. 10/6= .6
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10. When a rubber ball is dropped from a height, h cm on to a flat concrete floor, it rebounds to 3 5 h cm a height of Given a r
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Step-by-step explanation:

Let the height above which the ball is released be H

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An article suggested that yield strength (ksi) for A36 grade steel is normally distributed with μ = 42 and σ = 5.5.
alexandr1967 [171]

Answer:

a)P( X

We want this probability:

P( X >64)

And using the z score formula given by:

z = \frac{x -\mu}{\sigma}

We got:

P( X >64) =P(Z> \frac{64-42}{5.5}) =P(Z>4)=0.0000316

b) For this part we want to find a value a, such that we satisfy this condition:

P(X>a)=0.25   (a)

P(X   (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the 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=0.674. On this case P(Z<0.674)=0.75 and P(z>0.674)=0.25

If we use condition (b) from previous we have this:

P(X  

P(z

But we know which value of z satisfy the previous equation so then we can do this:

z=0.674

And if we solve for a we got

a=42 +0.674*5.5=45.707

So the value of height that separates the bottom 75% of data from the top 25% is 45.707.  

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 a

Let X the random variable that represent the heights of a population, and for this case we know the distribution for X is given by:

X \sim N(42,25.5)  

Where \mu=42 and \sigma=5.5

And we want this probability:

P( X

And using the z score formula given by:

z = \frac{x -\mu}{\sigma}

We got:

P( X

We want this probability:

P( X >64)

And using the z score formula given by:

z = \frac{x -\mu}{\sigma}

We got:

P( X >64) =P(Z> \frac{64-42}{5.5}) =P(Z>4)=0.0000316

Part b

For this part we want to find a value a, such that we satisfy this condition:

P(X>a)=0.25   (a)

P(X   (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the 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=0.674. On this case P(Z<0.674)=0.75 and P(z>0.674)=0.25

If we use condition (b) from previous we have this:

P(X  

P(z

But we know which value of z satisfy the previous equation so then we can do this:

z=0.674

And if we solve for a we got

a=42 +0.674*5.5=45.707

So the value of height that separates the bottom 75% of data from the top 25% is 45.707.  

8 0
3 years ago
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