Question

assume that when adults with smartphones are randomly selected 42 use them in meetings or classes if 15 adult smartphones are randomly selected, find the probability that "fewer" than 4 of them use their smartphones

Answer 1

Complete Question

Assume that when adults with smartphones are randomly selected 42% use them in meetings or classes if 15 adult smartphones are randomly selected, find the probability that "fewer" than 4 of them use their smartphones

Answer:

The  probability is  $P[X < 4]= 0.00314$

Step-by-step explanation:

From the question we are told that

      The proportion of those that use smartphone in meeting and classes is  p = 0.42

     The sample size is  $n = 15$

   The proportion of those that don't use  smartphone in meeting and classes is  

          $q = 1- p$

=>        $q = 1- 0.42$

=>         $q = 0.58$

Now  from the question we can deduce that the usage of the smartphone is having a  binomial  distribution since there is only two outcome  

So  the probability that "fewer" than 4 of them use their smartphones is mathematically evaluated as

       $P[X < 4 ] = P[X = 0] + P[X = 1] +P[X = 2] +P[X = 3]$=

=>     $P[X < 4] = [ \left 15} \atop {0}} \right. ] p^{15-0} * q^0 + [ \left 15} \atop {1}} \right. ] p^{15-1 }* q^1 + [ \left 15} \atop {2}} \right. ] p^{15-2 }* q^2 + [ \left 15} \atop {3}} \right. ] p^{15-3 }* q^3$

Where  $[\left 15} \atop {0}} \right. ]$ implies 15  combination 0 which has a value of  1 this is obtained using a scientific calculator

  So for the rest of the equation we will be making use of a scientific calculator to obtain the combinations

      $P[X < 4] = 1 * ^{15} * q^0 + 15 * p^{14 }* q^1 + 105 * p^{13 }* q^2 + 455 * p^{12 }* q^3$

substituting values

       $= 1 * (0.42)^{15} * (0.58)^0 + 15 * (0.42)^{14 }* (0.58)^1$

                     $+ 105 * (0.42)^{13 }* (0.58)^2 + 455 * (0.42)^{12 }* (0.58)^3$

=>       $P[X < 4]= 0.00314$