Respuesta :
Answer:
a) P(X ≤ 3) = 0.9946
P(X < 3) = 0.9639
b) P(X ≥ 4) = 0.0054
c) P(1 ≤ X ≤ 3) = 0.5313
d) E(X) = 0.75
σX = 0.84
e) P(X=0) = 0.0099
Step-by-step explanation:
We have x: number in the sample who have a food allergy. As the sample is of n=15 and p=0.05, we have:
[tex]X \sim Bin(15, 0.05)[/tex]
a) We have to determine P(X ≤ 3) and P(X < 3)
We can calculate P(X ≤ 3) as the sum of P(0), P(1), P(2) and P(3).
[tex]P(x\leq 3)=\sum_{k=0}^3P(k)\\\\\\P(x=0) = \binom{15}{0} p^{0}q^{15}=1*1*0.4633=0.4633\\\\P(x=1) = \binom{15}{1} p^{1}q^{14}=15*0.05*0.4877=0.3658\\\\P(x=2) = \binom{15}{2} p^{2}q^{13}=105*0.0025*0.5133=0.1348\\\\P(x=3) = \binom{15}{3} p^{3}q^{12}=455*0.0001*0.5404=0.0307\\\\\\P(x\leq 3)=0.4633+0.3658+0.1348+0.0307=0.9946[/tex]
P(x<3) can be calculated from the previos result as:
[tex]P(x<3)=P(X\leq3)-P(3)=0.9946-0.0307=0.9639[/tex]
b) We can calculate P(X ≥ 4) as:
[tex]P(X\geq4)=1-P(X<4)=1-P(X\leq3)=1-0.9946=0.0054[/tex]
c) We can calculate P(1 ≤ X ≤ 3) as:
[tex]P(1 \leq X \leq 3)=P(1)+P(2)+P(3)=0.3658+0.1348+0.0307=0.5313[/tex]
d) The expected value of a binomial variable is the product of the sample size n and the probability of success p:
[tex]E(X)=np=15*0.05=0.75[/tex]
The standard deviation is calculates as:
[tex]\sigma_x=\sqrt{np(1-p)}=\sqrt{15*0.05*0.95}=\sqrt{0.7125} =0.84[/tex]
e) In this case, the sample size is n=90.
We can calculate the probability that none has a food allergy as:
[tex]P(x=0) = \binom{90}{0} p^{0}q^{90}=0.95^{90}=0.0099[/tex]
