You are the coach of a basketball team that is currently looking for new players. One of the criteria for selection as a player is that the person must be above a particular height. Ideally, you want your next player to be as tall as possible. However, you do not want to rule out any potential players by making the cut-off height too strict. You decide that accepting players within the top 2.5% height bracket will be reasonable for your team. Assume that the height of all people follows a normal distribution with a mean of 70.3 in and a standard deviation of 2.1 in. Calculate the cut-off height (C) that ensures only people within the top 2.5% height bracket are allowed into the team. Give your answer in inches to the nearest inch.C = ______ in

Answer:

The  value is  [tex]x =  74.416 \ in[/tex]

Step-by-step explanation:

From the question we are told that

    The mean is  [tex]\mu  =   70.3\  in[/tex]

     The standard deviation is  [tex]\sigma  =  2.1\  in[/tex]

Generally the probability of  getting people with height  in the top 2.5% is mathematically represented as

  [tex]P(X >  x ) =  P(\frac{X - \mu }{\sigma}  > \frac{x - 70.3 }{2.1 }  ) = 0.025[/tex]

Generally    

         [tex]\frac{X - \mu }{\sigma}  =  Z (The  \  standardized \  value \  of X )[/tex]

=>    [tex]P(X >  x ) =  P(Z > \frac{x - 70.3 }{2.1 }  ) = 0.025[/tex]

Generally the critical  value of  0.025 from the normal distribution table is  

      [tex]Z_{0.025} =  1.96[/tex]

So

      [tex]\frac{x - 70.3 }{2.1 }   =  1.96[/tex]

=>   [tex]x =  74.416 \ in[/tex]

The cut-off height will be "74.416 in".

This same probability of an occurrence occurring is defined by probability. There are several real-life scenarios in something we should forecast the result of such an occurrence.

According to the question,

Mean, μ = 70.3 in

Standard deviation, σ = 2.1 in

Now,

The probability of getting people with height be:

→ P(X > x) = P([tex]\frac{X -\mu}{\sigma} > \frac{x-70.3}{2.1}[/tex])

                = 0.025

We know,

Z = [tex]\frac{X- \mu}{\sigma}[/tex]

→ P(X > x) = P(Z > [tex]\frac{x-70.3}{2.1}[/tex])

                = 0.025

The critical value be:

[tex]Z_{0.025}[/tex] = 1.96

hence,

The cut-off height be:

[tex]\frac{x-70.3}{2.1}[/tex] = 1.96

      x = 74.416 in

Thus the above answer is correct.

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