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Suppose 41% of the students in a university are baseball players. If a sample of 524 students is selected, what is the probability that the sample proportion of baseball players will be greater than 44%

Sagot :

Cricetus

Answer:

"0.0808" is the appropriate response.

Step-by-step explanation:

Given:

n = 524

[tex]\hat{P}[/tex] = 41%

or,

   = 0.41

[tex]1-\hat{P}=1-0.41[/tex]

         [tex]=0.59[/tex]

[tex]\mu \hat{P}=\hat{P}[/tex]

     [tex]=0.41[/tex]

Now,

⇒ [tex]6 \hat{P}=\sqrt{\frac{\hat {P}(1-\hat{P})}{n} }[/tex]

         [tex]=\sqrt{\frac{0.41\times 0.59}{524} }[/tex]

         [tex]=0.0215[/tex]

[tex]P(\hat {P}>44 \ percent)[/tex]

or,

[tex]P(\hat{P}>0.44)[/tex]

          [tex]=1-P(\hat{P}<0.44)[/tex]

          [tex]=1-P(\frac{\hat{P}-\mu \hat{P}}{6 \hat{P}} <\frac{0.44-0.41}{0.0215} )[/tex]

          [tex]=1-P(z<1.40)[/tex]

By using the standard normal table, we get

          [tex]=1-0.9192[/tex]

          [tex]=0.0808[/tex]    

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