Question

Consider a binomial experiment with 15 trials and probability 0.35 of success on a single trial.

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.