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shutvik [7]
4 years ago
14

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

Mathematics
1 answer:
svetoff [14.1K]4 years ago
3 0

Complete Question

The complete question is shown on the first uploaded image

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

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