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`

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`