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Afina-wow [57]
2 years ago
15

Major League Baseball now records information about every pitch thrown in every game of every season. Statistician Jim Albert co

mpiled data about every pitch thrown by 20 starting pitchers during the 2009 MLB season. The data set included the type of pitch thrown (curveball, changeup, slider, etc.) as well as the speed of the ball as it left the pitcher’s hand. A histogram of speeds for all 30,740 four-seam fastballs thrown by these pitchers during the 2009 season is shown below, from which we can see that the speeds of these fastballs follow a Normal model with mean μ = 92.12 mph and a standard deviation of σ = 2.43 mph.
Compute the z-score of pitch with speed 95.8 mph. (Round your answer to two decimal places.)



Approximately what fraction of these four-seam fastballs would you expect to have speeds between 91 mph and 93.2 mph? (Express your answer as a decimal, not a percent, and round to three decimal places.)



Approximately what fraction of these four-seam fastballs would you expect to have speeds below 91 mph? (Express your answer as a decimal, not a percent, and round to three decimal places.)



A baseball fan wishes to identify the four-seam fastballs among the fastest 20% of all such pitches. Above what speed must a four-seam fastball be in order to be included in the fastest 20%? (Round your answer to the nearest 0.1 mph.)
Mathematics
1 answer:
taurus [48]2 years ago
7 0
Fastballs i believe
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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
2 years ago
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