Question

In a large human population, cranial length is approximately normally distributed with a mean of 185.6 mm and a standard deviation of 12.7 mm. What is the probability that a random sample of size 10 from this population will have a mean less than than 180? a. 0.0174 b. 0.1357 c. 0.0823 d. -1.39

Answer 1

Answer:

The correct option is  c

Step-by-step explanation:

From the question we are told that

   The mean is  $\mu = 185.6 \ mm$

   The standard deviation is  $\sigma = 12.7 \ mm$

   The sample size is  n = 10  

    Generally the standard error of the mean is mathematically represented as

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

=>    $\sigma_{x} = \frac{ 12.7}{\sqrt{ 10 } }$

=>    $\sigma_{x} = 4.016$

Generally the probability that a random sample of size 10 from this population will have a mean less than than 180 is mathematically represented as

     $P(X < 180 ) = P(\frac{ X - \mu }{\sigma_{x} } < \frac{180 - 185.6}{4.016} )$

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

    $P(X < 180 ) = P(Z < - 1.394 )$

From the z table  the area under the normal curve to the left corresponding to    -1.394  is  

          $P(X < 180 ) = P(Z < - 1.394 ) = 0.081659$

=>      $P(X < 180 ) = 0.0823$