Question

The heights of North American women are nor-mally distributed with a mean of 64 inches and a standard deviation of 2 inches. a. b. c. What is the probability that a randomly selected woman is taller than 66 inches? A random sample of four women is selected. What is the probability that the sample mean height is greater than 66 inches? What is the probability that the mean height of a random sample of 100 women is greater than

Answer 1

Complete Question

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Answer:

a

  $P(X > 66) = P(Z> 1 ) = 0.15866$

b

$P(\= X > 66) = P(Z> 2 ) = 0.02275$

c

$P(\= X > 66) = P(Z> 10 ) = 0$

Step-by-step explanation:

From the question we are told that

   The  population mean is  $\mu = 64 \ inches$

    The standard deviation is  $\sigma = 2 \ inches$

The probability that a randomly selected woman is taller than 66 inches   is mathematically represented as

    $P(X > 66) = P(\frac{X - \mu }{\sigma } > \frac{ 66 - \mu }{\sigma} )$

Generally $\frac{ X - \mu }{\sigma } = Z(The \ standardized \ value \ of \ X )$

So  

      $P(X > 66) = P(Z> \frac{ 66 - 64 }{ 2} )$

     $P(X > 66) = P(Z> 1 )$

From the z-table  the value of  $P(Z > 1 ) = 0.15866$

 So  

       $P(X > 66) = P(Z> 1 ) = 0.15866$

Considering b

 sample mean is  n  =  4  

Generally the standard error of mean is mathematically represented as

        $\sigma _{\= x} = \frac{\sigma }{\sqrt{4} }$

=>    $\sigma _{\= x} = \frac{2 }{\sqrt{4} }$

=>    $\sigma _{\= x} = 1$

The probability that the sample mean height is greater than 66 inches    

     $P(\= X > 66) = P(\frac{X - \mu }{\sigma_{\= x } } > \frac{ 66 - \mu }{\sigma_{\= x }} )$

=>   $P(\= X > 66) = P(Z > \frac{ 66 - 64 }{1} )$

=>  $P(\= X > 66) = P(Z> 2 )$

From the z-table  the value of  $P(Z > 2 ) = 0.02275$

=> $P(\= X > 66) = P(Z> 2 ) = 0.02275$

Considering b

 sample mean is  n  =  100

Generally the standard error of mean is mathematically represented as

        $\sigma _{\= x} = \frac{2 }{\sqrt{100} }$

=>    $\sigma _{\= x} = 0.2$

The probability that the sample mean height is greater than 66 inches

   $P(\= X > 66) = P(Z > \frac{ 66 - 64 }{0.2} )$

=>  $P(\= X > 66) = P(Z> 10 )$

From the z-table  the value of  $P(Z > 10 ) = 0$

$P(\= X > 66) = P(Z> 10 ) = 0$