Question

Consider a binomial experiment with 15 trials and probability 0.35 of success on a single trial.
(a) Use the binomial distribution to find the probability of exactly 10 successes (Round your answer to three decimal places.)
(b) Use the normal distribution to approximate the probability of exactly 10 successes. (Round your answer to three decimal places.) (
(c) Compare the results of parts (a) and (b).
A. These results are fairly different.
B. These results are almost exactly the same.

Answer 1

Answer:

a

   $P(X = 10 ) = 0.0096$

b

   $P(X = 10 ) = 0.0085$

c

 Option A is correct

Step-by-step explanation:

From the question we are told that

     The sample size is   n =  15

     The  probability of success is  $p = 0.35$

     The number of success we are considering is  r = 10  

 

Now the probability of failure is mathematically evaluated as

        $q = 1- p$

substituting value

       $q = 1- 0.35$

       $q = 0.65$

Now using  the binomial distribution  to find the probability of exactly 10 successes we have that

    $P(X = r ) = [\left n } \atop {r}} \right. ] * p^r * q^{n- r}$

substituting values

    $P(X = 10 ) = [\left 15 } \atop {10}} \right. ] * p^{10}* q^{15- 10}$

Where  $[\left 15 } \atop {10}} \right. ]$ mean 15 combination 10  which is evaluated with a calculator to obtain  

       $[\left 15 } \atop {10}} \right. ] = 3003$

So

      $P(X = 10 ) = 3003 * 0.35 ^{10}* 0.65^{15- 10}$

       $P(X = 10 ) = 0.0096$

Now using  the normal distribution to approximate the probability of exactly 10 successes, we have that

  $P(X = r ) = P( r < X < r )$

Applying continuity correction

          $P(X = r ) = P( r -0.5 < X < r +0.5)$

substituting values

        $P(X = 10) = P( 10-0.5 < X < 10+0.5)$

       $P(X = 10 ) = P( 9.5 < X < 10.5)$

Standardizing  

         $P(X = r ) = P( \frac{9.5 - \mu }{\sigma } < \frac{X - \mu }{\sigma } < \frac{10.5 - \mu}{\sigma } )$

The  where  $\mu$ is the mean which is mathematically represented as

        $\mu = n * p$

substituting values

        $\mu = 15 * 0.35$

         $\mu = 5.25$

The standard deviation is evaluated as      

     $\sigma = \sqrt{n * p * q }$

substituting values

    $\sigma = \sqrt{15 * 0.35 * 0.65 }$

    $\sigma = 1.8473$

Thus  

     $P(X = 10 ) = P( \frac{9.5 - 5.25 }{1.8473 } < \frac{X - 5.25 }{1.8473 } < \frac{10.5 - 5.25}{1.8473 } )$

     $P(X = 10 ) = P( 2.30 < Z < 2.842 )$

     $P(X = 10 ) = P(Z < 2.842 ) - P(Z < 2.30 )$

From the normal distribution table we obtain the $P(Z < 2.841)$ as

      $P(Z < 2.841) = 0.99775$

And  the  $P(Z < 2.30)$

     $P(Z < 2.30) = 0.98928$

There value can also be obtained from a probability of z calculator at (Calculator dot net website)

So  

    $P(X = 10) = 0.99775 - 0.98928$

     $P(X = 10 ) = 0.0085$

Looking at the calculated values for question a and b  we see that the values are fairly different.