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krek1111 [17]
3 years ago
10

The average weight an NFL player is 245.7 pounds with standard deviation of 34.5 pounds but probability distribution of the popu

lation is unknown. If a random sample of 32 NFL players is selected, what is the probability that:
(i) the average weight of the sample will be less than 234 pounds?

(ii) the average weight of the sample is between 248-254?

(iii) the average weight of the sample is between 242-251?
Mathematics
1 answer:
Dvinal [7]3 years ago
6 0
Answer: For part I, the probability is 36.69% that the average weight will be less than 234.

To find the percents for each one, you need to find the z-score and use a normal distribution table. A z-score is the number of standard deviations above or below the mean.

For 234, the z-score would be:  (234 - 245.7) / 34.5 = -0.34 which corresponds to a percent of 36.69%.

For that last 2 parts, you follow the same plan. However, you will find two different percents. You will have to subtract them to find the percent between them.

For Part II, you should get 6.85%

For Part III, you should get 10.37%.
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maw [93]

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The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

( − ∞ , ∞ )  

{x | x  ∈  R }


Hope this helps I'm 80% sure this right.



4 0
3 years ago
I need help on thi and the first person who answer correctly gets a BRANLIST​ and show work​
const2013 [10]

Answer:

w<6 THE SMALL LINE UNDER THE LESS THAN SIGN IS STILL THERE I JUST CANNOT ADD IT

Step-by-step explanation:

first solve the equation to the right.

-3(2w+1) = -6w-3

now solve the whole equation

-33-w< -6w-3

-w+6w < -3+33

5w < 30

divide the 5 from both sides to simplify

5/5w < 30/5

w < 6

THE ANSWER IS w < 6

3 0
3 years ago
An electrical rm manufactures light bulbs that have a life span that is approximately normally distributed. The population stand
stira [4]

Answer:

The 't' test statistic = 1.46 < 1.69

The test of hypothesis is H 0 :μ = 800 is accepted

A sample of 30 bulbs are found came from average µ= 800

Step-by-step explanation:

Step 1:-

Given population of mean μ = 800

given size of small sample n =30

sample standard deviation 'S' = 45

Mean value of the sample χ = 788

Null hypothesis H_{0} =µ  =800

alternative hypothesis  H_{1} = µ ≠ 800

<u>Step 2</u>:-

The 't' test statistic t =  \frac{x-μ}{\frac{sig}{\sqrt{n} } }

                             t = \frac{788-800}{\frac{45}{\sqrt{30} } }

                       t = \frac{12}{8.215} = 1.4607

<u>Step 3</u>:-

The degrees of freedom γ = n-1 = 30-1 =29

From "t" value from table at 0.05 level of significance ( t = 1.69)

The calculated value t = 1.4607 < 1.69

Therefore The null hypothesis H_{0} ' is accepted.

<u>conclusion</u>:-

A sample of 30 bulbs are came found from average µ= 800

7 0
4 years ago
A motorcycle cost $12,000 when it was purchased. The value of a motorcycle decreases by 6% each year. Find the rate of decay eac
dalvyx [7]
<span>the rate of decay each month can calculated using the following formula:
</span><span>
6%/year × 1 year/(12 months) = 0.5%/month
0.06 x 1/12 = 0.005/month
</span><span>So the 0.5% is the right answer.
I hope it helped.</span>
4 0
3 years ago
An animal shelter takes in an average of 5 animals per day. The shelter must keep its total occupancy below 300. Currently, the
Akimi4 [234]

Answer:

B. x<27

Step-by-step explanation:

300-165=135

135/5=27

8 0
4 years ago
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