According to the National Institute of Allergy and Infectious Diseases, 6% of American adults have a food allergy. A large company plans a lunch reception for its 500 employees. Assume that employees are independent. Let the random variable X be the number of company employees who have a food allergy. (a) What are the assumptions/requirements of a Binomial distribution? Does this situation meet all these requirements?
(b) What are the expected value and standard deviation of X (i.e., of the population)?
(c) What is the probability that none of the 500 employees has a food allergy?

Answers

Answer 1

Answer:

Answer:

a) The requirements is that each trial can only have two outcomes(success/failure), and each trial has the same probability of a success, that is, they are independent of each other.

In this problem, for each person, either they are allergic to some kind of food, or they are not. The probability of a person being allergic is independent of any other people. So this situations meets all these requirements.

b) The expected value is 30 and the standard deviation is 5.31.

c) Close to 0% probability that none of the 500 employees has a food allergy

Step-by-step explanation:

For each adult, there are only two possible outcomes. Either they have food allergy, or they do not. The probability of an adult having allergy is independent of other adults. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)$

In which $C_(n,x)$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_(n,x) = (n!)/(x!(n-x)!)$

And p is the probability of X happening.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$√(V(X)) = √(np(1-p))$

6% of American adults have a food allergy

This means that $p = 0.06$

500 employees.

This means that $n = 500$

(a) What are the assumptions/requirements of a Binomial distribution? Does this situation meet all these requirements?

The requirements is that each trial can only have two outcomes(success/failure), and each trial has the same probability of a success, that is, they are independent of each other.

(b) What are the expected value and standard deviation of X (i.e., of the population)?

$E(X) = np = 500*0.06 = 30$

$√(V(X)) = √(np(1-p)) = √(500*0.06*0.94) = 5.31$

The expected value is 30 and the standard deviation is 5.31.

(c) What is the probability that none of the 500 employees has a food allergy?

This is P(X = 0).

$P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)$

$P(X = 0) = C_(500,0).(0.06)^(0).(0.94)^(500) \cong 0$

Close to 0% probability that none of the 500 employees has a food allergy

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On his last 5 math tests, Gilbert earned: 94, 72, 83, 94, 90. What was his average score? ( Report your answer to the tenths place).

Answers

Answer:

Average of Gilbert in last 5 test = 86.6

Step-by-step explanation:

Given:

Score earned by Gilbert in last 5 test = 94, 72, 83, 94, 90

Find:

Average of Gilbert in last 5 test

Computation:

Average value = Sum of all events / Total number of events

Average of Gilbert in last 5 test = Sum of all last results / Number of results

Average of Gilbert in last 5 test = [94 + 72 + 83 + 94 + 90] / 5

Average of Gilbert in last 5 test = [433] / 5

Average of Gilbert in last 5 test = 86.6

10 ) Which Property of Multiplication is shown? (5 + 3) x 2 = 5 x 2 + 3 x 2A. Identity Property
C. Associative Property
B. Distributive Property
D. Commutative Property

Answers

Property of Multiplication that is shown on the expression (5 + 3) x 2 = 5 x 2 + 3 x 2 is Distributive Property (option b)

The property of multiplication that is shown in the given equation is the Distributive Property. The Distributive Property states that multiplication distributes over addition. In other words, when we multiply a number by the sum of two other numbers, it is the same as multiplying the number by each of the individual addends and then adding the results together.

In the given equation, we have "(5 + 3) x 2 = 5 x 2 + 3 x 2." Let's break down the left-hand side first. "(5 + 3)" represents the sum of 5 and 3, which is 8. So, we have "8 x 2." By applying the Distributive Property, we can rewrite this as "5 x 2 + 3 x 2."

Now, on the right-hand side, we have "5 x 2 + 3 x 2." Here, we are multiplying 5 by 2 and 3 by 2 individually and then adding the products together. This aligns with the Distributive Property, which states that the result will be the same as on the left-hand side.

So, the Main Answer is option B. Distributive Property

To know more about multiplication here

brainly.com/question/29488963

#SPJ3

Answer:

B. Distributive Property of Multiplication

Step-by-step explanation:

The given equation shows that two similar expressions are equal to each other.

How? Well, they both equal 16 to start. These are the same expressions - one is just distributed out!

$2 \cdot (5 + 3) \n2 \cdot 5 + 2 \cdot3 \n10 + 6 = 16$

and...

$5 \cdot 2 + 3 \cdot 2\n10 + 6\n16$

In the equations, the only difference is that we distributive 2 and multiplied in by 5 and 3!

I hope I was of assistance! #SpreadTheLove <3

10) Solve the inequality to find how many hours until the temperature willbe below 32 degrees F
A)3 hours
B)3.5 hours
C)4 hours
D)4.5 hours
PLS ANSWER

Answers

Answer:

D)4.5 hours

hope this helps

plz mark brainleist

Please help me I’ve been stuck for hours

Answers

Answer/Step-by-step explanation:

Given that a and b are two corresponding sides of two similar figures, it follows that the ratio of their areas = a²/b².

We would use the above knowledge to solve the questions given as follows:

✔️The first pair similar of shapes:

Missing area = x cm²

Therefore,

x/9 = 8²/4²

x/9 = 64/16

x/9 = 4

Cross multiply

x = 4 × 9

x = 36 cm²

✔️The second pair similar of shapes:

Missing area = x cm²

Therefore,

x/240 = 8²/32²

x/240 = 64/1,024

Cross multiply

x*1,024 = 64*240

x*1,024 = 15,360

Divide both sides by 1,024

x = 15,360/1,024

x = 15 cm²

✔️The third pair similar of shapes:

Missing area = x cm²

Therefore,

x/40 = 3²/2²

x/40 = 9/4

Cross multiply

x*4 = 9*40

x*4 = 360

Divide both sides by 4

x = 360/4

x = 90 cm²

In tilapia, an important freshwater food fish from Africa, the males actively court females. They have more incentive to court a female who has already laid all of her eggs, but can they tell the difference? an experiment was done to measure the male tilapia's response to the smell of female fish. Water containing feces from females that were either pre-ovulatory (they still had eggs) or post-ovulatory (they had already laid their eggs) was washed over the gills of males hooked up to an electro-olfactogram machine which measured when the senses of the males were excited. The amplitude of the electro-olfactogram was used as a measure of the excitability of the males in the two different circumstances. Six males were exposed to the scent of pre-ovulatory females; their readings average 1.51 with a standard deviation of .25. Six different males were exposed to post-ovulatory females; their average readings of 0.87 with standard deviation is .31. Assume that the electro-olfactogram readings were approximately normally distributed within the groups.(A) test for a difference in the excitability of the males with exposure to these two types of females
(B) what is the estimated average difference in electro-olfactogram readings between the two groups? What is the 95% confidnece limit for the difference between population means?

Answers

Answer:

a) $t=\frac{1.51-0.87}{\sqrt{(0.25^2)/(6)+(0.31^2)/(6)}}=3.936$

"=T.INV(1-0.025,10)", and we got $t_(critical)=\pm 2.28$

Statistical decision

Since our calculated value is higher than our critical value, $z_(calc)=3.936>2.28=t_(critical)$ , we have enough evidence to reject the null hypothesis at 5% of significance.

b) $(\bar X_1 -\bar X_2) \pm t_(\alpha/2)\sqrt{(s^2_(1))/(n_(1))+(s^2_(2))/(n_(2))}$

The degrees of freedom are given:

$df = n_1 + n_2 -2 = 6+6-2 = 10$

$(1.51 -0.87) - 2.28\sqrt{(0.25^2)/(6)+(0.31^2)/(6)}= 0.269$

$(1.51 -0.87) + 2.28\sqrt{(0.25^2)/(6)+(0.31^2)/(6)}= 1.010$

Step-by-step explanation:

Part a

Data given and notation

$\bar X_(1)=1.51$ represent the mean for scent of pre ovulatory

$\bar X_(2)=0.87$ represent the mean for post ovolatory

$s_(1)=0.25$ represent the sample standard deviation for preovulatory

$s_(2)=0.31$ represent the sample standard deviation for postovulatory

$n_(1)=6$ sample size for the group preovulatory

$n_(2)=6$ sample size for the group postovulatory

z would represent the statistic (variable of interest)

$p_v$ represent the p value

Concepts and formulas to use

We need to conduct a hypothesis in order to check if the mean's are different, the system of hypothesis would be:

H0: $\mu_(1) = \mu_(2)$

H1: $\mu_(1) \neq \mu_(2)$

If we analyze the size for the samples both are lower than 30, so for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{\bar X_(1)-\bar X_(2)}{\sqrt{(s^2_(1))/(n_(1))+(s^2_(2))/(n_(2))}}$ (1)

z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

Calculate the statistic

We have all in order to replace in formula (1) like this:

$t=\frac{1.51-0.87}{\sqrt{(0.25^2)/(6)+(0.31^2)/(6)}}=3.936$

Find the critical value

We find the degrees of freedom:

$df = n_1 + n_2 -2 = 6+6-2 = 10$

In order to find the critical value we need to take in count that we are conducting a two tailed test, so we are looking for thwo values on the t distribution with df =10 that accumulates 0.025 of the area on each tail. We can us excel or a table to find it, for example the code in Excel is:

"=T.INV(1-0.025,10)", and we got $t_(critical)=\pm 2.28$

Statistical decision

Since our calculated value is higher than our critical value, $z_(calc)=3.936>2.28=t_(critical)$ , we have enough evidence to reject the null hypothesis at 5% of significance.

Part b

For this case the confidence interval is given by:

$(\bar X_1 -\bar X_2) \pm t_(\alpha/2)\sqrt{(s^2_(1))/(n_(1))+(s^2_(2))/(n_(2))}$

The degrees of freedom are given:

$df = n_1 + n_2 -2 = 6+6-2 = 10$

$(1.51 -0.87) - 2.28\sqrt{(0.25^2)/(6)+(0.31^2)/(6)}= 0.269$

$(1.51 -0.87) + 2.28\sqrt{(0.25^2)/(6)+(0.31^2)/(6)}= 1.010$

List all the factor pairs for 48 make a tabletop to help