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ANEK [815]
2 years ago
9

A college football coach has decided to recruit only the heaviest 15% of high school football players. He knows that high school

players’ weights are normally distributed and that this year, the mean weight is 225 pounds with a standard deviation of 43 pounds. Calculate the weight at which the coach should start recruiting players.
Mathematics
1 answer:
Alla [95]2 years ago
3 0

Answer:

The coach should start recruiting players with weight 269.55 pounds or more.                                                    

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 225 pounds

Standard Deviation, σ = 43 pounds

We are given that the distribution of weights is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

We have to find the value of x such that the probability is 0.15

P( X > x) = P( z > \displaystyle\frac{x - 225}{43})=0.15  

= 1 -P( z \leq \displaystyle\frac{x - 225}{43})=0.15  

=P( z \leq \displaystyle\frac{x - 225}{43})=0.85  

Calculation the value from standard normal z table, we have,  

P(z < 1.036) = 0.85

\displaystyle\frac{x - 225}{43} = 1.036\\\\x = 269.548 \approx 269.55

Thus, the coach should start recruiting players with weight 269.55 pounds or more.

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Step-by-step explanation:

Let x = number of books that Hannah donated

We know that the number of students is 150 and the total number of books is represented by b, we can write that as:

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Now we know that Hannah donated 3 times as much as the average. We can write that as:

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Tesla wanted to determine the average miles per kWh that their vehicles get across all models and variations. They took a sample
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Answer:

\mu_{\bar x} = \mu = E(X) =30KWh

\sigma_{\bar X}= \frac{\sigma}{\sqrt{n}}=\frac{3}{\sqrt{100}}=0.3

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".  

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Solution to the problem

For this case we select a sample of n =100

From the central limit theorem we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

So then the sample mean would be:

\mu_{\bar x} = \mu = E(X) =30KWh

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