Question

NBC News reported on May 2, 2013, that 1 in 20 children in the United States have a food allergy of some sort. Consider selecting a random sample of 15 children and let X be the number in the sample who have a food allergy. Then X Bins, 0.05). (Round your probabilities to three decimal places) (a) Determine both PX S 3) and Px 3) (b) Determine P(X 2 4) (c) Determine Pi s x 3). (d) what are E(a) and ax? (Round your aswers to twe decimal places) (e) In@sample of 90 children, what is the probability that none has food allergy?

Answer 1

Answer:

Part a: The value of P(X≤3) is 0.995 while that of P(X<3) is 0.964

Part b: The value of P(X≥4) is 0.005

Part c: The value of P(1≤X≤3) is 0.531

Part d: The expected value of mean is 0.75 while that of standard deviation is 0.844.

Part e: The probability that none has allergy in 90 children in 0.010.

Step-by-step explanation:

Part a

For the value is given as

$P(X\leq 3)=P(0\leq X\leq 3)\\P(X\leq 3)=P(X=0)+P(X=1)+P(X=2)+P(X=3)$

$P(X=x)=^{n}C_{x}(p)^{x}(1-p)^{n-x}$

Here p=0.05, n=15 so the values are given as

$P(X=0)=^{15}C_{0}(0.05)^{0}(1-0.05)^{15-0}=^{15}C_{0}(0.05)^{0}(0.95)^{15}\approx 0.4633$

$P(X=1)=^{15}C_{1}(0.05)^{1}(1-0.05)^{15-1}=^{15}C_{1}(0.05)^{1}(0.95)^{14}\approx 0.3657$

$P(X=2)=^{15}C_{2}(0.05)^{2}(1-0.05)^{15-2}=^{15}C_{2}(0.05)^{2}(0.95)^{13}\approx 0.1348$

$P(X=3)=^{15}C_{3}(0.05)^{3}(1-0.05)^{15-3}=^{15}C_{3}(0.05)^{3}(0.95)^{12}\approx 0.0307$

So

$P(X\leq 3)=P(0\leq X\leq 3)\\P(X\leq 3)=P(X=0)+P(X=1)+P(X=2)+P(X=3)\\P(X\leq 3)=0.4633+0.3657+0.1348+0.0307=0.99452 \approx 0.995$

So the value of P(X≤3) is 0.995

For the value is given as

$P(X< 3)=P(0\leq X <3)\\P(X< 3)=P(X=0)+P(X=1)+P(X=2)$

$P(X=x)=^{n}C_{x}(p)^{x}(1-p)^{n-x}$

Here p=0.05, n=15 so the values are given as

$P(X=0)=^{15}C_{0}(0.05)^{0}(1-0.05)^{15-0}=^{15}C_{0}(0.05)^{0}(0.95)^{15}\approx 0.4633$

$P(X=1)=^{15}C_{1}(0.05)^{1}(1-0.05)^{15-1}=^{15}C_{1}(0.05)^{1}(0.95)^{14}\approx 0.3657$

$P(X=2)=^{15}C_{2}(0.05)^{2}(1-0.05)^{15-2}=^{15}C_{2}(0.05)^{2}(0.95)^{13}\approx 0.1348$

So

$P(X< 3)=P(0\leq X<3)\\P(X<3)=P(X=0)+P(X=1)+P(X=2)\\P(X<3)=0.4633+0.3657+0.1348=0.96379 \approx 0.964$

So the value of P(X<3) is 0.964

Part b

For the value is given as

$P(X\geq 4)=1-P(X< 4)=1-P(0\leq X< 4)\\P(X\geq 4)=1-[P(X=0)+P(X=1)+P(X=2)+P(X=3)]$

$P(X=x)=^{n}C_{x}(p)^{x}(1-p)^{n-x}$

Here p=0.05, n=15 so the values are given as

$P(X=0)=^{15}C_{0}(0.05)^{0}(1-0.05)^{15-0}=^{15}C_{0}(0.05)^{0}(0.95)^{15}\approx 0.4633$

$P(X=1)=^{15}C_{1}(0.05)^{1}(1-0.05)^{15-1}=^{15}C_{1}(0.05)^{1}(0.95)^{14}\approx 0.3657$

$P(X=2)=^{15}C_{2}(0.05)^{2}(1-0.05)^{15-2}=^{15}C_{2}(0.05)^{2}(0.95)^{13}\approx 0.1348$

$P(X=3)=^{15}C_{3}(0.05)^{3}(1-0.05)^{15-3}=^{15}C_{3}(0.05)^{3}(0.95)^{12}\approx 0.0307$

So

$P(X\geq 4)=1-P(X=0)+P(X=1)+P(X=2)+P(X=3)\\P(X\geq 4)=1-[0.4633+0.3657+0.1348+0.0307P(X\leq 3)]=1-0.99452=0.00548 \approx 0.005$

So the value of P(X≥4) is 0.005

Part c

For the value is given as

$P(1\leq X\leq 3)\\P(1\leq X\leq 3)=P(X=1)+P(X=2)+P(X=3)$

$P(X=x)=^{n}C_{x}(p)^{x}(1-p)^{n-x}$

Here p=0.05, n=15 so the values are given as

$P(X=1)=^{15}C_{1}(0.05)^{1}(1-0.05)^{15-1}=^{15}C_{1}(0.05)^{1}(0.95)^{14}\approx 0.3657$

$P(X=2)=^{15}C_{2}(0.05)^{2}(1-0.05)^{15-2}=^{15}C_{2}(0.05)^{2}(0.95)^{13}\approx 0.1348$

$P(X=3)=^{15}C_{3}(0.05)^{3}(1-0.05)^{15-3}=^{15}C_{3}(0.05)^{3}(0.95)^{12}\approx 0.0307$

So

$P(1\leq X\leq 3)=P(X=1)+P(X=2)+P(X=3)\\P(1\leq X\leq 3)=0.3657+0.1348+0.0307=0.531241 \approx 0.531$

So the value of P(1≤X≤3) is 0.531

Part d

The expected value of mean is given as

$\mu=np$

Here n =15, p =0.05

$\mu=np\\\mu=15*0.05\\\mu=0.75$

The expected value of mean is 0.75

While the expected value of the standard deviation is given as

$\sigma=\sqrt{np(1-p)}$

Here n =15, p =0.05

$\sigma=\sqrt{np(1-p)}\\\sigma=\sqrt{15*0.05(1-0.05)}\\\sigma=\sqrt{15*0.05(0.95)}\\\sigma=0.844$

So the expected value of standard deviation is 0.844

Part e

For the value is given as

$P(X= 0)$

$P(X=x)=^{n}C_{x}(p)^{x}(1-p)^{n-x}$

Here p=0.05, n=90 so the values are given as

$P(X=0)=^{90}C_{0}(0.05)^{0}(1-0.05)^{90-0}=^{90}C_{0}(0.05)^{0}(0.95)^{90}\approx 0.010$

So the probability that none has allergy in 90 children in 0.010.