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 ~ Bin(15, 0.05). (Round your probabilities to three decimal places.)
(a)
Determine both P(X ≤ 3) and P(X < 3).
P(X ≤ 3)
=
P(X < 3)
=
(b)
Determine P(X ≥ 4).
P(X ≥ 4)
=
(c)
Determine P(1 ≤ X ≤ 3).
P(1 ≤ X ≤ 3) =
(d)
What are E(X) and σ_{X}? (Round your answers to two decimal places.)
E(X)
=
σ_{X}
=
(e)
In a sample of 70 children, what is the probability that none has a food allergy?
You may need to use the appropriate table in the Appendix of Tables to answer this question.
a)
Here, n = 15, p = 0.05, (1 - p) = 0.95 and x = 3
As per binomial distribution formula P(X = x) = nCx * p^x * (1 -
p)^(n - x)
We need to calculate P(X <= 3).
P(X <= 3) = (15C0 * 0.05^0 * 0.95^15) + (15C1 * 0.05^1 *
0.95^14) + (15C2 * 0.05^2 * 0.95^13) + (15C3 * 0.05^3 *
0.95^12)
P(X <= 3) = 0.4633 + 0.3658 + 0.1348 + 0.0307
P(X <= 3) = 0.995
Here, n = 15, p = 0.05, (1 - p) = 0.95 and x = 3
As per binomial distribution formula P(X = x) = nCx * p^x * (1 -
p)^(n - x)
We need to calculate P(X < 3).
P(X < 3) = (15C0 * 0.05^0 * 0.95^15) + (15C1 * 0.05^1 * 0.95^14)
+ (15C2 * 0.05^2 * 0.95^13)
P(X < 3) = 0.4633 + 0.3658 + 0.1348
P(X < 3) =0.964
2)
Here, n = 15, p = 0.05, (1 - p) = 0.95 and x = 4
As per binomial distribution formula P(X = x) = nCx * p^x * (1 -
p)^(n - x)
We need to calculate P(X >= 4).
P(X >= 4) = (15C4 * 0.05^4 * 0.95^11) + (15C5 * 0.05^5 *
0.95^10) + (15C6 * 0.05^6 * 0.95^9) + (15C7 * 0.05^7 * 0.95^8) +
(15C8 * 0.05^8 * 0.95^7) + (15C9 * 0.05^9 * 0.95^6) + (15C10 *
0.05^10 * 0.95^5)
P(X >= 4) = 0.005 + 0.001 + 0 + 0 + 0 + 0 + 0
P(X >= 4) = 0.006
c)
Here, n = 15, p = 0.05, (1 - p) = 0.95, x1 = 1 and x2 = 3.
As per binomial distribution formula P(X = x) = nCx * p^x * (1 -
p)^(n - x)
We need to calculate P(1 <= X <= 3)
P(1 <= X <= 3) = (15C1 * 0.05^1 * 0.95^14) + (15C2 * 0.05^2 *
0.95^13) + (15C3 * 0.05^3 * 0.95^12)
P(1 <= X <= 3) = 0.366 + 0.135 + 0.031
P(1 <= X <= 3) = 0.532
d)
mean = np = 15 * 0.05 = 0.75
Std.dev = sqrt(npq)
= sqrt(15 * 0.05 * 0.95)
= 0.84
e)
Here, n = 70, p = 0.05, (1 - p) = 0.95 and x = 0
As per binomial distribution formula P(X = x) = nCx * p^x * (1 -
p)^(n - x)
We need to calculate P(X = 0)
P(X = 0) = 70C0 * 0.05^0 * 0.95^70
P(X = 0) = 0.028
NBC News reported on May 2, 2013, that 1 in 20 children in the United States...
NBC News reported on May 2, 2013, thot 1 in 20 children in the United States have e 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 Bi1,.). (Round your probabilities to three decimal places.) (a) Determine both P(X S 3) and P(X 3) P(X s 3) (b) Determine P(X 4) A(X24) . () Determine P[1 sXS 3), What are...
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 25 children and let X be the number in the sample who have a food allergy. Then X ~ Bin(25, .05). a. Determine both P(X <= 3) and P(X < 3). b. Determine P(X >= 4). c. Determine P(1 <= X <= 3). d. What are E(X) and Sigma (x)? e....
16. + -18 points DevoreStat9 3.E.048. My Notes Ask Your Teacher 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 25 children and let X be the number in the sample who have a food allergy. Then X - Bin(25, 0.05). (Round your probabilities to three decimal places.) mais des contre cele de tornare in the under the server con...
When circuit boards used in the manufacture of compact disc players are tested, the long-run percentage of defectives is 5%. Let X = the number of defective boards in a random sample of size n = 25, so X~ Bin(25, 0.05). (Round your probabilities to three decimal places.) (a) Determine P(X ≤ 2). (b) Determine P(X ≥ 5). (c) Determine P(1≤ X≤ 4). (d) What is the probability that none of the 25 boards is defective? (e) Calculate the expected value and standard deviation of X. (Round...
According to the U.S. Census Bureau, 11% of children in the United States lived with at least one grandparent in 2009 (USA TODAY, June 30, 2011). Suppose that in a recent sample of 1560 children, 249 were found to be living with at least one grandparent. At a 1% significance level, can you conclude that the proportion of all children in the United States who currently live with at least one grandparent is higher than 11%? Use both the p-value...
--- According to the U.S. Census Bureau, 11% of children in the United States lived with at least one grandparent in 2009 (USA TODAY, June 30, 2011). Suppose that in a recent sample of 1570 children, 236 were found to be living with at least one grandparent. At a 2.5% significance level, can you conclude that the proportion of all children in the United States who currently live with at least one grandparent is higher than 11%? Use both the...
The Statistical Abstract of the United States published by the U.S. Census Bureau reports that the average annual consumption of fresh fruit per person is 99.9 pounds. The standard deviation of fresh fruit consumption is about 30 pounds. Suppose a researcher took a random sample of 38 people and had them keep a record of the fresh fruit they ate for one year. Appendix A Statistical Tables (Round all z values to 2 decimal places. Round your answers to 4...
Problem 3 Suppose that 20,000 married adults in the United States were as theoretical probabilities. Let X equal the number of children, with PDF as per the table below. All entries are to be decimal values accurate to the nearest 0.01. surveyed as to the number of children they have. The results are compiled and used P) P() 0 01 1 005 2 015 3 03 0.15 6 01 Complete the table above a. b. Determine the expected value. EX...
Suppose that 20, 000 married adults in the United States were surveyed as to the number of children they have. The results are compiled and used as theoretical probabilities. Let X equal the number of children, with PDF as per the table below. All entries are to be decimal values accurate to the nearest 0.01 P(x) x. P(x) Collapse Al 0.15 2 0.15 3 0.3 0.2 5 0.05 0.1 a. Complete the table above. b. Determine the expected value. EX]...
n at least According to th Humane Society of the United States, there are approximately 77.5 million o ned dogs in the United States, and approximately 40% of all US households o one d09-t Suppose that the 40% figure is correct and that 15 households are randomly selected for a pet ownership survey. (a) What is the probablity that exactly eight of the households have at least one dog? (Round your answer to three decimal places.) (b) What is the...