Respuesta :
Answer:
a) P(X<3)=0.964
P(X≤3)=0.995
b) P(X≥4)=0.005
P(1≤X≤3)=0.532
d) E(X)=0.75
Sigma=0.056
e) P(x=0)=0.077
Step-by-step explanation:
We have a binomial distribution with parameters n=15 and p=0.05.
The probabiltiy that k children, out of a sample of 15 children, have a food allergy can be calculated as:
[tex]P(x=k) = \dbinom{n}{k} p^{k}(1-p)^{n-k}\\\\\\P(x=k) = \dbinom{15}{k} 0.05^{k} 0.95^{15-k}\\\\\\[/tex]
a) P(X<3) can be calculated as the sum written as:
[tex]P(x<3)=P(x=0)+P(x=1)+P(x=2)\\\\\\P(x=0) = \dbinom{15}{0} p^{0}(1-p)^{15}=1*1*0.463=0.463\\\\\\P(x=1) = \dbinom{15}{1} p^{1}(1-p)^{14}=15*0.05*0.488=0.366\\\\\\P(x=2) = \dbinom{15}{2} p^{2}(1-p)^{13}=105*0.0025*0.513=0.135\\\\\\\\P(x<3)=0.463+0.366+0.135=0.964[/tex]
We can use this result to calculate P(X≤3) as:
[tex]P(x\leq3)=P(x<3)+P(x=3)\\\\\\P(x=3) = \dbinom{15}{3} p^{3}(1-p)^{12}=455*0.0001*0.54=0.031\\\\\\\\P(x\leq3)=0.964+0.031=0.995[/tex]
b) We can calculate P(X≥4) as:
[tex]P(X\geq4)=1-P(X<4)=1-P(x\leq3)\\\\P(X\geq4)=1-0.995=0.005[/tex]
We can calculate P(1≤X≤3) as:
[tex]P(1\leq x\leq3)=P(x=1)+P(x=2)+P(x=3)\\\\P(1\leq x\leq3)=0.366+0.135+0.031=0.532[/tex]
c) The mean and standard deviation can be calcalated as:
[tex]E(x)=n\cdot p=15\cdot0.05=0.75\\\\\sigma=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.05\cdot0.95}{15}}=\sqrt{0.00317}=0.056[/tex]
d) We can calculate this as the probability of a child not having an alergy, multiplied 50 times:
[tex]P(x=0)=(1-p)^{n}=0.95^{50}=0.077[/tex]
