Question

The mean of a population is 74 and the standard deviation is 16. The shape of the population is unknown. Determine the probability of each of the following occurring from this population. Appendix A Statistical Tables a. A random sample of size 36 yielding a sample mean of 75 or more b. A random sample of size 140 yielding a sample mean of between 72 and 75 c. A random sample of size 217 yielding a sample mean of less than 74.7

Answer 1

Answer:

a

 $P(X > 75)= 0.35402$

b

$P(72 < X < 75 ) = 0.2529$

c

$P( X < 74.7) = 0.74041$

Step-by-step explanation:

From the question we are told that

  The population mean is  $\mu = 74$

  The population standard deviation is  $\sigma = 16$

 Considering question a  

    The sample size is  n  =  36  

Generally the standard error of mean is mathematically represented as

     $\sigma_{x} = \frac{\sigma }{\sqrt{n} }$

=>  $\sigma_{x} = \frac{16}{\sqrt{36} }$

=>  $\sigma_{x} = 2.67$

Generally the probability that a  random sample of size 36 yielding a sample mean of 75 or more is mathematically represented as

     $P(X > 75) = P( \frac{X - \mu }{ \sigma_{x}} > \frac{75 - 74}{ 2.67 } )$

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

   $P(X > 75) = P(Z > 0.3745 )$

From the z table  the area under the normal curve representing 0.3745 to the right is  

     $P(Z > 0.3745 ) = 0.35402$

=>   $P(X > 75)= 0.35402$

 Considering question b  

    The sample size is  n  =  104

Generally the standard error of mean is mathematically represented as

     $\sigma_{x} = \frac{\sigma }{\sqrt{n} }$

=>  $\sigma_{x} = \frac{16}{\sqrt{104} }$

=>  $\sigma_{x} = 1.5689$

Generally the probability that a random sample of size 104 yielding a sample mean  between 72 and 75 is mathematically represented as

      $P(72 < X < 75 ) = P(\frac{72 - 74 }{1.5689} < \frac{X - \mu }{\sigma_{x}} < \frac{75 - 74 }{1.5689} )$

=>   $P(72 < X < 75 ) = P(-1.275 < Z < 0.375 )$

=>   $P(72 < X < 75 ) = P(Z < 0.375 ) - P(Z < -1.275)$

From the z table  the area under the normal curve representing -1.275 to to the left is

   $P(Z < -1.275) =0.10115$

=> $P(72 < X < 75 ) = 0.35402 - 0.10115$

=> $P(72 < X < 75 ) = 0.2529$

Considering question c

    The sample size is  n  =  217

Generally the standard error of mean is mathematically represented as

     $\sigma_{x} = \frac{\sigma }{\sqrt{n} }$

=>  $\sigma_{x} = \frac{16}{\sqrt{217} }$

=>  $\sigma_{x} = 1.086$

Generally the probability that a  random sample of size 217 yielding a sample mean of less than 74.7 is mathematically represented as

       $P( X < 74.7) = P(\frac{X - \mu }{\sigma_x} < \frac{ 74.7 - 74 }{ 1.086 })$

=>   $P( X < 74.7) = P(Z < 0.6446 )$

From the z table  the area under the normal curve representing 0.6446  to to the left is

     $P(Z < 0.6446 ) = 0.74041$

=>  $P( X < 74.7) = 0.74041$