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 [tex]P[X < 4]= 0.00314[/tex]
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 [tex]n = 15[/tex]
The proportion of those that don't use smartphone in meeting and classes is
[tex]q = 1- p[/tex]
=> [tex]q = 1- 0.42[/tex]
=> [tex]q = 0.58[/tex]
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
[tex]P[X < 4 ] = P[X = 0] + P[X = 1] +P[X = 2] +P[X = 3][/tex]=
=> [tex]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[/tex]
Where [tex][\left 15} \atop {0}} \right. ][/tex] 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
[tex]P[X < 4] = 1 * ^{15} * q^0 + 15 * p^{14 }* q^1 + 105 * p^{13 }* q^2 + 455 * p^{12 }* q^3[/tex]
substituting values
[tex]= 1 * (0.42)^{15} * (0.58)^0 + 15 * (0.42)^{14 }* (0.58)^1[/tex]
[tex]+ 105 * (0.42)^{13 }* (0.58)^2 + 455 * (0.42)^{12 }* (0.58)^3[/tex]
=> [tex]P[X < 4]= 0.00314[/tex]
