Answer:
a) 83.69%
b) 84.97%
c) 4.74%
Step-by-step explanation:
This situation can be modeled with the Binomial Distribution which figures out the likelihood of an event that occurs exactly k times out of n, and is given by
[tex]\large P(k;n)=\binom{n}{k}p^kq^{n-k}[/tex]
where
[tex]\large \binom{n}{k}[/tex]= combination of n elements taken k at a time.
p = probability that the event (“success”) occurs once
q = 1-p
In this case, the event “success” is grossing over $850 one day with a probability of 70% = 0.7
a)
“Estimate the probability that the store will gross over $850 at least 3 out of 5 business days”
Here we want to find
P(3;5)+P(4;5)+P(5;5), that is to say,
[tex]\large \binom{5}{3}0.7^30.3^2+\binom{5}{4}0.7^40.3+\binom{5}{5}0.7^50.3^0=0.3087+0.36015+0.16807 =0.83692 =83.69\%[/tex]
b)
“At least 6 out of 10 business days”
Now we are looking for
P(6;10)+P(7;10)+P(8;10)+P(9;10)+P(10;10)
Applying the same formula with n=10, we get
=0.20012 + 0.26683+0.23347+0.12106+0.82825=0.84973 = 84.97%
c)
“Fewer than 5 out of 10 business days”
P(0;10)+P(1;10)+P(2;10)+P(3;10)+P(4;10) =
[tex]\large 5.9049*10^{-6}+0.0001378+0.001447+0.009+0.03676 = 0.04735 = 4.74\%[/tex]
