MATHEMATICS COLLEGE

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 σX? (Round your answers to two decimal places.) E(X) = σX = (e) In a sample of 90 children, what is the probability that none has a food allergy?

Answers

Answer 1

Answer:

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:

$X \sim Bin(15, 0.05)$

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).

$P(x\leq 3)=\sum_(k=0)^3P(k)\n\n\nP(x=0) = \binom{15}{0} p^(0)q^(15)=1*1*0.4633=0.4633\n\nP(x=1) = \binom{15}{1} p^(1)q^(14)=15*0.05*0.4877=0.3658\n\nP(x=2) = \binom{15}{2} p^(2)q^(13)=105*0.0025*0.5133=0.1348\n\nP(x=3) = \binom{15}{3} p^(3)q^(12)=455*0.0001*0.5404=0.0307\n\n\nP(x\leq 3)=0.4633+0.3658+0.1348+0.0307=0.9946$

P(x<3) can be calculated from the previos result as:

$P(x<3)=P(X\leq3)-P(3)=0.9946-0.0307=0.9639$

b) We can calculate P(X ≥ 4) as:

$P(X\geq4)=1-P(X<4)=1-P(X\leq3)=1-0.9946=0.0054$

c) We can calculate P(1 ≤ X ≤ 3) as:

$P(1 \leq X \leq 3)=P(1)+P(2)+P(3)=0.3658+0.1348+0.0307=0.5313$

d) The expected value of a binomial variable is the product of the sample size n and the probability of success p:

$E(X)=np=15*0.05=0.75$

The standard deviation is calculates as:

$\sigma_x=√(np(1-p))=√(15*0.05*0.95)=√(0.7125) =0.84$

e) In this case, the sample size is n=90.

We can calculate the probability that none has a food allergy as:

$P(x=0) = \binom{90}{0} p^(0)q^(90)=0.95^(90)=0.0099$

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Which of the following values cannot be probabilities? 0.04, 5 divided by 3, 1, 0, 3 divided by 5, StartRoot 2 EndRoot, negative 0.59, 1.49 Select all the values that cannot be probabilities. A. 1.49 B. 1 C. three fifths D. StartRoot 2 EndRoot E. five thirds F. 0 G. negative 0.59 H. 0.04

Answers

Answer:

A. 1.49

D. √2

E. five thirds

G. - 0.59

Step-by-step explanation:

In order to be a probability, a value must be at least zero, or at most 1:

$0 \leq P\leq 1$

Evaluating each of the given values:

A. 1.49

1.49 is at least zero but it is greater than one, therefore 1.49 cannot be a probability.

B. 1

1 represents a probability of 100%, therefore this value can be a probability

C. three fifths

$0\leq (3)/(5) \leq 1$

Can be a probability

D. √2

$\sqrt 2 =1.41 > 1$

Cannot be a probability

E. five thirds

$(5)/(3)=1.67>1$

Cannot be a probability

F. 0

0 represents a probability of 0%, therefore this value can be a probability

G. - 0.59

Negative values cannot be probabilities.

H. 0.04

$0\leq 0.04 \leq 1$

Can be a probability

Final answer:

Probabilities are values ranging from 0 to 1, inclusive. With this in mind, values 5/3, √2, -0.59, and 1.49 cannot be probabilities as they're either below 0 or above 1.

Explanation:

In the field of mathematics, specifically in statistics, a probability represents the likelihood of an event occurring and is always a value between 0 and 1, inclusively. The value 0 means that an event will not happen, whilst 1 means the event is certain to happen. Therefore, any value less than 0 or greater than 1 cannot be a probability.

Given the values: 0.04, 5 divided by 3, 1, 0, 3 divided by 5, √2, negative 0.59, and 1.49, the values that cannot be probabilities are:

Value 5 divided by 3 (which equals approximately 1.67)
Value √2 (which equals approximately 1.41)
Negative 0.59
1.49

These numbers do not lie within the range of 0 to 1, and hence, cannot represent probabilities.

Learn more about Probability here:

brainly.com/question/22962752

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How many points separated the gold medalist and the fifth place competitor?

Answers

Answer:

10 point separation.

Step-by-step explanation:

There are 14 points for gold and 4 points for fifth place.

Hope this helps!

Answer:

!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Step-by-step explanation:

26.67

The difference between a number x and 12 is 16?

Answers

Answer:

x equals 28

Step-by-step explanation:

16 + 12 = x because

What is the greatest common factor of 12 and 90?

Answers

The first step to find the gcf of 12 and 90 is to list the factors of each number. The factors of 12 are 1, 2, 3, 4, 6 and 12. The factors of 90 are 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45 and 90. So the greatest common factor is 90 Buddy.

Whoever answers correctly gets brainlist

Answers

Answer:

So now D would be at (-4 , 2)

Step-by-step explanation:

From D, you can calculate every other letter’s location. Yeah.

Answer:

Please mark me as brainliest!

Step-by-step explanation:

The coordinates:

D: ( 2,6 )

A: ( 2, 1 )

B: ( 5, 1 )

C: ( 5, 6 )

You want to translate this down 4 units and left 6 units.

So what you would do is this:

Formula = ( x - 6 , y - 4 )

D: ( -4, 2 )

A: ( -4, -3 )

B: ( -1, -3 )

C: ( -1, 2 )

Complete the pattern 4,15,7,12,10,9 ? ? ? show me how you got your work

Answers

Given the pattern:
4,15,7,12,10,9,..
Every other term after the first term is increasing by 3. Thus continuing the pattern we shall have:
4, 15, 7, 12, 10, 9, 13, 11,16
thus the missing numbers will be:
13,11,16

Split this pattern into two sub-patterns.

We can see that odd position terms first, third, and fifth terms are increasing by 3.

We also can see that even position terms second, fourth, and sixth terms are decreasing by 3.

So the remaining term is seventh, eighth, and ninth terms. So from pattern above, seventh and ninth is continue after odd position term, so that would be 13 then 16.

eighth term is continue after even position term. That would be 6.

So then the remaining pattern would be 13, 6, 16.

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