Answer:

Pr(at most 2 are defective) = 0.38

Step-by-step explanation:

probability of defective = 0.05

probability of not defective =1- 0.05 = 0.95

Pr(at most 2 are defective) = pr (0 defective) + pr (1 defective) + pr (2defective)

pr (0 defective) = (0.05^0) * (0.95^20) = 0.36

pr (1 defective) = (0.05^1) * (0.95^19) = 0.019

pr (2 defective) = (0.05^2) * (0.95^18) = 0.0025 * 0.397 = 0.00099

Pr(at most 2 are defective) = 0.36 + 0.019 + 0.00099

Pr(at most 2 are defective) = 0.37999 = 0.38

Answer:

P(x2)=1- P(x

A family has j children with probability pj , where p1 = .1, p2 = .25, p3 = .35, p4 = .3. A child from this family is randomly chosen. Given that the child is the eldest child in the family, find the conditional probability that the family has

Answer:

A) the conditional probability that the has only 1 child = 0.24

B) The conditional probability that the family has only 4 children = 0.18

Step-by-step explanation:

To answer the questions, we first start with defining each event. Let E be the event that the child selected is the oldest and let Fj be the event that the family has j children.

From this, we can deduce that the probability that the child is the oldest, given that there is j children is ; P(E | Fj ) = 1/j.

In addition, we know P(Fj ) = pj as given in the problem. In answering the 2 questions, we seek the probability P(Fj | E). Thus, by the Bayes’s formula;

P(Fj | E) = P(EFj )/P(E) which gives;

P(Fj | E) = {P(E | Fj )P(Fj )} / {Σ(4,i=1)P(E | Fj )P(Fj )}

= (1/j)/ Σ(4,i=1)(1/j)pj

= ((pj)/j) / {p1 + (p2)/2 + (p3)/3 +(p4)/4}

Therefore, the conditional probability that the family has only 1 child; P(F1 | E) = p1 / {p1 + (p2)/2 + (p3)/3 +(p4)/4} = 0.1/ (0.1 + (0.25/2) +(0.35/3) + (0.3/4) = 0.1/0.4167 = 0.24

The conditional probability that the family has only 4 children =

{(p4)/4} / {p1 + (p2)/2 + (p3)/3 +(p4)/4} = (0.3/4)/ (0.1 + (0.25/2) +(0.35/3) + (0.3/4) = 0.075/0.4167 = 0.18

An environmental group at a local college is conducting independent tests to determine the distance a particular make of automobile will travel while consuming only 1 gallon of gas. They test a sample of five cars and obtain a mean of 28.2 miles. Assuming that the standard deviation is 2.7 miles, find the 95 percent confidence interval for the mean distance traveled by all such cars using 1 gallon of gas.

Answer:

Step-by-step explanation:

Confidence Interval

When the population standard deviation is known, the formula for a confidence interval for a population mean is:

Where n is the sample size and z is the corresponding z-value from the standard normal distribution for the selected confidence level. The value of z for a 95% confidence interval is z=1.96. The rest of the values are

Calculating the confidence interval

Or, equivalently

Can y = sin(t2) be a solution on an interval containing t = 0 of an equation y + p(t) y + q(t) y = 0 with continuous coefficients? Explain your answer.

Answer:

Step-by-step explanation:

y = sin(t^2)

y' = 2tcos(t^2)

y'' = 2cos(t^2) - 4t^2sin(t^2)

so the equation become

2cos(t^2) - 4t^2sin(t^2) + p(t)(2tcos(t^2)) + q(t)sin(t^2) = 0

when t=0, above eqution is 2. That is, there does not exist the solution. so y can not be a solution on I containing t=0.

387 in base 10 to base 5

Answer: 387 in base 10 to base 5 is 3022

Step-by-step explanation:

To convert 387 in base 10 to base 5, we would take the following steps

Firstly , we would divide the number to be converted by 5. The remainder forms the last digit of the number in base 5 while the quotient is divided again to get a new quotient and remainder. The new remainder forms the next digit to the left of the last digit. It continues till it gets to zero. Therefore,

387/5 = 77 remainder 2(last digit)

77/5 = 15 remainder 2(next digit to the left of the last).

15/5 = 3 remainder 0(next digit to the left)

3/5 = 0 remainder 3(next and final digit to the left)

Therefore, 387 in base 10 to base 5 is 3022

Answer:

3022

Step-by-step explanation:

To convert a number from base 10 to base 5, divide the number repeatedly by 5, keeping track of each remainder, until you get a quotient that is equal to zero

Next, write all the remainders in reverse order.

387₁₀ = 3022₅

Check:

387 = 3×5³ + 0×5² + 2×5¹ + 2×5⁰

387 = 3×125 + 0×25 + 2×5 + 2×1

387 = 375 + 0 + 10 + 2

387 = 387

OK.

It has been conjectured by the U.S. Census Bureau that "approximately 60% of foreign-born people who live in the U.S. are not naturalized citizens." In a national random sample of 70 foreign-born people who live in the U.S., on average, how many people would you expect to get that are not naturalized citizens? Select the best answer below.a. 28 peopleb. 42 peoplec. 4.10 peopled. None of these.

Answer:

Option B) 42 people

Step-by-step explanation:

We are given the following in the question:

Percentage of people who live in the U.S. that are not naturalized citizens = 60%

Sample size, n = 70

We have to given an estimate for people who are not naturalized citizens.

Thus, 42 people are not naturalized citizens.

Option B) 42 people

Find the area under the standard normal distribution curve for the following intervals. a. Between z = 0 and z = 2.0 b. To the right of z = 1.5 c. To the left of z = −1.75 d. Between z = −2.78 and z = 1.66

Answer:

a)

And we can use the following excel code to find the probability:

"=NORM.DIST(2,0,1,TRUE)-NORM.DIST(0,0,1,TRUE)"

b) 1.5) =1-P(Z" alt=" P(Z>1.5) =1-P(Z" align="absmiddle" class="latex-formula">

And we can use the following code and we got:

"=1-NORM.DIST(1.5,0,1,TRUE)"

1.5) =1-P(Z" alt=" P(Z>1.5) =1-P(Z" align="absmiddle" class="latex-formula">

c)

And we can use the following code and we got:

"=NORM.DIST(-1.75,0,1,TRUE)"

d)

And we can use the following excel code to find the probability:

"=NORM.DIST(1.66,0,1,TRUE)-NORM.DIST(-2.78,0,1,TRUE)"

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Part a

We want this probability:

And we can use the following excel code to find the probability:

"=NORM.DIST(2,0,1,TRUE)-NORM.DIST(0,0,1,TRUE)"

Part b

For this case we want this probability:

1.5)" alt=" P(Z>1.5)" align="absmiddle" class="latex-formula">

And we can use the complement rule and we have:

1.5) =1-P(Z" alt=" P(Z>1.5) =1-P(Z" align="absmiddle" class="latex-formula">

And we can use the following code and we got:

"=1-NORM.DIST(1.5,0,1,TRUE)"

1.5) =1-P(Z" alt=" P(Z>1.5) =1-P(Z" align="absmiddle" class="latex-formula">

Part c

We want this probability:

And we can use the following code and we got:

"=NORM.DIST(-1.75,0,1,TRUE)"

Part d

We want this probability:

And we can use the following excel code to find the probability:

"=NORM.DIST(1.66,0,1,TRUE)-NORM.DIST(-2.78,0,1,TRUE)"

The area under the standard normal distribution curve for the interval between z = 0 and z = 2.0 is; 0.47725

How to use the normal distribution table?A) P(0 < z < 2)

From online p-value from two z-score calculator, we have;

p-value = (0.97725 - 0.5000) = 0.47725

B) P(z > 1.5)

From online p-value from z-score calculator, we have;

p-value = 0.0668

C) P(z < 1.75)

From online p-value from z-score calculator, we have;

p-value = 1 - 0.040059

p-value = 0.9599

D) P(-2.78 < z < 1.6)

From online p-value from two z-score calculator, we have;

p-value = 0.94882

Read more about Normal Distribution table at; brainly.com/question/4079902

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Find four square roots of 2833 modulo 4189. (The modulus factors as 4189 = 59 · 71. Note that your four square roots should be distinct modulo 4189.)Hoffstein, Jeffrey. An Introduction to Mathematical Cryptography (Undergraduate Texts in Mathematics) (p. 112). Springer New York. Kindle Edition.

Answer:

Step-by-step explanation:

Let us proceed to find square roots of modulo 59 and 71:

≡ 2833 mod 59 ≡ 1 mod 59 (1)

≡ 2833 mod 71 ≡ 64 mod 71 (2)

By inspection, we find that = ± 1 and = ± 8 works

Now, using Chinese remainder to solve the simultaneous congruence,

≡

The first congruence yields

Then putting this back into the second equation, we get

≡ ⇒ ≡ ⇒ ≡

But

≡ ;

Hence,

≡ ⇒ ≡

This shows that is a third square root. From this, we immediately get the fourth square root, namely ≡ .

Note that the square roots:

are all distinct modulo 4189.

Find the probability that a randomly selected card from a Euchre deck is a jack or a spade. (Enter the fraction reduced to the lowest possible fraction on the left hand side of "=" symbol and the final answer on the right hand side simplified to the lowest possible fraction.)

Answer:

3/8

Step-by-step explanation:

There are 24 cards in a euchre deck with 4 Jack and 6 spade

P (J U S) = 4/24 + 6/24 - 1/24 = 1/6 + 1/4 - 1/24 = 3/8

Find the minimum sample size necessary to be 99% confident that the population mean is within 3 units of the sample mean given that the population standard deviation is 29.

Answer:

The minimum sample size required is 207.

Step-by-step explanation:

The (1 - α) % confidence interval for population mean μ is:

The margin of error of this confidence interval is:

Given:

*Use a z-table for the critical value.

Compute the value of n as follows:

Thus, the minimum sample size required is 207.

Find the probability that a randomly selected multiple birth for women 15-54 years old involved a mother who was at least 40 years old. P(at least 40

Age Number of Multiple Births

15–19 100

20–24 467

25–29 1620

30–34 2262

35–39 1545

40–44 328

45–54 105

---------------------------

Total. 6427

-----------------------------

Answer:

Probability of selecting a mother who is at least 40 years old is 0.067.

That means there are 6.7% of the the total mothers who are at least 40 years old.

Step-by-step explanation:

The question is looking for probability of a mother at least 40 years.

It means 40 years and above.

From the table, the number of mothers who fall into this category = 328 +105 = 433

So, the probability would be = 433/6427 = 0.067

Therefore, converting to percentage, we have 6.7%

Answer:3/8

Step-by-step explanation:

Total number of women with ages from 15 to 54years (15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54)= 40

Total number of women with age 40 years and above = 15

Probability of picking a mother of age greater than 40 = 15/40 which is equal to 3/8.

Find a possible exponential function in y = a · bx (y=a(b)^x) form that represents the situation described below. Has an initial value of 2 and passes through the point (3, 128).

Answer:

y = 2 · 4ˣ

Step-by-step explanation:

Initial value of 2 means that a = 2.

y = 2 · bˣ

It passes through the point (3, 128), so:

128 = 2 · b³

64 = b³

b = 4

y = 2 · 4ˣ

Karen, 28 years old and a single taxpayer, has a salary of $33,000 and rental income of $33,000 for the 2019 calendar tax year. Karen is covered by a pension through her employer. AGI phase-out range for traditional IRA contributions for a single taxpayer who is an active plan participant is $64,000 – $74,000. a. What is the maximum amount that Karen may deduct for contributions to her traditional IRA for 2019?

Answer:

$4,800

Step-by-step explanation:

The maximum contribution for traditional IRA in 2019 = $6000

Given that;

karen has a salary of $33,000 and rental income of $33,000; then total income = $66,000

AGI phase-out range for traditional IRA contributions for a single taxpayer who is an active plan participant is $64,000 – $74,000.

PhaseOut can be calculated as:

=

= 0.2 * 6000

= 1200

Therefore, the maximum amount that Karen may deduct for contributions to her traditional IRA for 2019 = The maximum contribution for traditional IRA in 2019 - PhaseOut

= $6000 - $1,200

= $4,800

Let's say you take an exam worth 240 points and your score marked the 78th percentile of all the exam scores. Based on this you can conclude that ______% of exam takers had scores equal to or better than yours.

Here, we are required to determine what percentage of exam takers had scores equal to or better than mine.

The correct answer is therefore, 22%

First, a percentile is a statistical term used to represent each of the 100 equal groups into which a population can be divided according to the distribution of values of a particular variable, (in this case the scores).

Therefore, according to the definition above, Since my score marked the 78th percentile, we can conclude that only (100 - 78)% of the exam takers had scores equal to or better than mine.

The correct answer is therefore, 22%

Read more:

brainly.com/question/15876171

Answer:

28%

Step-by-step explanation:

78th percentile means 78% scored less than/equal to you.

100 - 78 = 22%

how strong is the relationship between the score on the first exam and the score on teh final exam in an elementary statistics course? ehre are data for eight students form such a course

Answer: Invalid question

Step-by-step explanation: Hello friend, kindly make available the remaining details of the question.

Regards

How many zeros are at the end of 458 · 885? Explain how you can answer this question without actually computing the number. (Hint: 10 = 2 · 5.) When this number is written in ordinary decimal form, each 0 at its end comes from a factor of

Answer:

1

Step-by-step explanation:

458 = 2 × 229

885 = 5 × 177

Factors 2 and 5 are appearing only once, hence it's a multiple of 10.

Which means 1 zero at the end

One card is selected at random from an ordinary deck of 52 playing cards. Events A, B, and C are defined below. Find the probabilities for parts (a) through (h) below and express your results in words. Compute the conditional probabilities directly; do not use the conditional probability rule. Note that the ace has the highest value. A equals= event a card higher than 9card higher than 9 is selected B equals= event a card between 7 and 10 comma inclusive commacard between 7 and 10, inclusive, is selected C equals= event a spadespade is selected

Answer:

P(A) = 29/52

P (B)= 8/52

Probability of spade = P(C)= 13/52

Step-by-step explanation:

A equals= event a card higher than 9 is selected

B equals= event a card between 7 and 10

C equals= event a spade is selected

A "standard" deck of playing cards consists of 52 Cards in each of the 4 suits of Spades, Hearts, Diamonds, and Clubs. Each suit contains 13 cards: Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King.

If we consider " that the ace has the highest value" than the cards higher than 9 would be 8,7,6,5,4,3,2. Seven cards for each suit and there are four suits and an Ace also

So the P(A) = 29/52

For B we know there are two numbers between 7 and 10 that are 8 and 9.

So for four suits that would be 2*4= 8 cards

Probability of B = P(B)= 8/52

For C there are 13 spades so Probability of spade = P(C)= 13/52

One large course has 900 students, broken down into section meetings with 30 students each. The section meetings are led by TA's. On the final, the class avg is 63 and the SD is 20. However, in one section the avg. is only 55. The TA argues this way:If you took 30 students at random from the class, there is a good chance they would avg below 55 on the final. Chance variation.Is this a good argument? Yes or no, explain.

Answer:

The argument made by the TA is not good.

Step-by-step explanation:

The argument made by the TA is that in a random sample of 30 students there is a good chance that the class average would be below 55.

As the sample size is large (n ≥ 30), the sampling distribution of sample mean will follow a Normal distribution.

The mean and standard deviation of this sampling distribution are:

The information provided is:

Compute the probability that sample mean is less than 55 as follows:

The probability that a random sample of 30 students have a class average below 55 is 0.0143.

The probability is very small.

The event of 30 student having an average below 55 is an unusual event.

Unusual events are those events that have a very low probability of occurrence.

Thus, the argument made by the TA is not good.

Following is a sample of the number of words spoken in each inauguration address for U.S. presidents. 1125, 2978, 1172, 1507, 4776, 3801, 2015, 2463, 4388, 135, 3967, 1,620 , 1437, 3217, 1340, 2308, 1681, 2546, 2158, 2170, 5433, 1729, 698 What is the 32nd percentile? Write only a number as your answer.

Answer:

Step-by-step explanation:

Positional or index value of mth= m/100 × total number of observations.

32/100 ×23= 7.36th term

Arrange the data in ascending order:

135, 698,1125, 1172,1340, 1437, 1507, 1620, 1681, 1729, 2015, 2158,2170, 2308, 2463, 2546, 2978, 3217, 3801, 3967, 4388, 4776, 5433.

The 7th term is 1507

Therefore, the 32nd percentile is 1507

Answer: 1620

Step-by-step explanation: first we arrange the values in increasing order

135 698 1125 1172 1340 1437 1507 1620 1681 1729 2015 2158 2170 2308 2463 2546 2978 3217 3801 3967 4388 4776 5433

The number of numbers is 23

To calculate the rank of values that falls below the 32nd percentile we use

R = p/100 * N

R is the rank

P is the percentile 32

N is the number of numbers 23

Therefore

R = 32/100 *23 = 7.36

we approximate to get 7

This means that the numbers up to the 7th number is below the 32nd percentile

That mean the answer is the 8th number 1620

Alternatively we can simply use the formular

R = p/100 * (N + 1)

R = 32/100 * (23 +1) = 7.68

Approximately 8

So the 8th value is the answer and that is 1620

Scores on the Critical Reading part of the SAT exam in a recent year were roughly Normal with mean 495 and standard deviation 118. You choose an SRS of 100 students and average their SAT Critical Reading scores. If you do this many times, the standard deviation of the average scores you get will be close to (a) 118 (b) 118/100=1.18. (c) 118/V100 11.8

Answer:

(c) 118/V100 11.8

Step-by-step explanation:

We use the central limit theorem to solve this question.

The Central Limit Theorem estabilishes that, for a random variable X, with mean and standard deviation , a large sample size can be approximated to a normal distribution with mean and standard deviation

In this problem, we have that:

So

So the correct answer is:

(c) 118/V100 11.8

Suppose you can somehow choose two people at random who took the SAT in 2014. A reminder that scores were Normally distributed with mean and stanard deviation of 1497 and 322, respectively. What is the probability that both of them scored above a 1520? Assume that the scores of the two test takers are independent.

The required probability is

Independent probability:An event can be called an independent of another event if the probability of occurrence of one event is not affected by the occurrence of the other. Suppose two cards are drawn one after the other.

Let be the scores of the exam.

Now, solving 1520)" alt="P(X > 1520)" align="absmiddle" class="latex-formula">

1520)=P(\frac{x-\mu}{\sigma} > \frac{1520-1497}{322} )\\=P(z > 0.0714)\\=1-0.5285\\=0.4715" alt="P(X > 1520)=P(\frac{x-\mu}{\sigma} > \frac{1520-1497}{322} )\\=P(z > 0.0714)\\=1-0.5285\\=0.4715" align="absmiddle" class="latex-formula">

Now, let and be the scores of those two people where and are given to be independent.

Then,

1520,x_2 > 1520)=P(x_1 > 1520)\times P(x_2 > 1520)\\=0.4715 \times 0.4715\\=0.2223" alt="P(x_1 > 1520,x_2 > 1520)=P(x_1 > 1520)\times P(x_2 > 1520)\\=0.4715 \times 0.4715\\=0.2223" align="absmiddle" class="latex-formula">

Learn more about the topic of Independent probability:

brainly.com/question/26169642

Answer:

22.29% probability that both of them scored above a 1520

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean and standard deviation , the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

The first step to solve the question is find the probability that a student has of scoring above 1520, which is 1 subtracted by the pvalue of Z when X = 1520.

So

has a pvalue of 0.5279

1 - 0.5279 = 0.4721

Each students has a 0.4721 probability of scoring above 1520.

What is the probability that both of them scored above a 1520?

Each students has a 0.4721 probability of scoring above 1520. So

22.29% probability that both of them scored above a 1520

A particle moves along the parabola y equals x squared in the first quadrant in such a way that its x-coordinate (measured in meters) increases at a steady 10 StartFraction m Over sec EndFraction . How fast is the angle of inclination theta of the line joining the particle to the origin changing when x equals 1 m question mark
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