MATHEMATICS COLLEGE

In a study of government financial aid for college students, it becomes necessary to estimate the percentage of full-time college students who earn a bachelor's degree in four years or less. Find the sample size needed to estimate that percentage. Use a 0.02 margin of error and use a confidence level of 99%. Complete parts (a) through (c) below.a. Assume that nothing is known about the percentage to be estimated.n = __b. Assume prior studies have shown that about 55% of full-time students earn bachelor's degrees in four years or less.n = _c. Does the added knowledge in part (b) have much of an effect on the sample size?

Answers

Answer 1

Answer:

Answer:

(a) The sample size required is 2401.

(b) The sample size required is 2377.

Step-by-step explanation:

The confidence interval for population proportion p is:

$CI=\hat p\pm z_(\alpha/2)\sqrt{(\hatp(1-\hat p))/(n)}$

The margin of error in this interval is:

$MOE=z_(\alpha/2)\sqrt{(\hatp(1-\hat p))/(n)}$

The information provided is:

MOE = 0.02

$z_(\alpha/2)=z_(0.05/2)=z_(0.025)=1.96$

(a)

Assume that the proportion value is 0.50.

Compute the value of n as follows:

$MOE=z_(\alpha/2)\sqrt{(\hat p(1-\hat p))/(n)}\n0.02=1.96* \sqrt{(0.50(1-0.50))/(n)}\nn=(1.96^(2)*0.50(1-0.50))/(0.02^(2))\n=2401$

Thus, the sample size required is 2401.

(b)

Given that the proportion value is 0.55.

Compute the value of n as follows:

$MOE=z_(\alpha/2)\sqrt{(\hat p(1-\hat p))/(n)}\n0.02=1.96* \sqrt{(0.55(1-0.55))/(n)}\nn=(1.96^(2)*0.55(1-0.55))/(0.02^(2))\n=2376.99\n\approx2377$

Thus, the sample size required is 2377.

(c)

On increasing the proportion value the sample size decreased.

14 units

Step-by-step explanation:

The distance between two points can be found by using the formula

$d = \sqrt{ ({x1 - x2})^(2) + ({y1 - y2})^(2) } \n$

where

(x1 , y1) and (x2 , y2) are the points

From the question the points are

(3, 5.25) and (3, –8.75)

The distance between them is

$d = \sqrt{ ({3 - 3})^(2) + ({5.25 + 8.75})^(2) } \n = \sqrt{ {0}^(2) + {14}^(2) } \n = √(196) \: \: \: \: \: \: \: \: \: \: \n = 14 \: \: \: \: \: \:$

We have the final answer as

14 units

Hope this helps you

30 points!!! Help me on 1. 2. 3.

Answers

I think 1 is to press C

Explain how to use the distributive property to find an expression that is equivalent to 20+10

Answers

Answer: 2(10+5)

Step-by-step explanation: half of 20 is 10. half of 10 is five and i put them both in parentheses. outside the parenthesis i placed 2 so it is multiplied to the inner set of numbers.

Correct answer- 2(10+5) :)

A particular concentration of a chemical found in polluted water has been found to be lethal to 40% of the crayfish that are exposed to the concentration for 24 hours. 22 crayfish are placed in a tank containing this concentration of chemical in the water. (a) What is the probability that 13 or 17 survive?
(b) What is the probability that at least 17 survive?
(c) What is the probability that at most 16 survive?
(d) What number of crayfish are expected to survive?
(e) What is the variance in the number of crayfish that are expected to survive? No actual crayfish were harmed in the making of this question.

Answers

Answer:

a) $P(X=13)=(24C13)(0.4)^(13) (1-0.4)^(24-13)=0.0608$

$P(X=17)=(24C17)(0.4)^(17) (1-0.4)^(24-17)=0.0017$

$P(X=13 U X=17) = 0.0608+0.0017=0.0624$

b) $P(X \geq 17)=0.000427$

c) $P(X \geq 16)= 1-0.000427=0.9996$

d) $E(X)= np = 22*0.4=8.8$

e) $Var(X) = np(1-p)= 22*0.4*(1-0,4)=5.28$

Step-by-step explanation:

1) Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

2) Solution to the problem

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=24, p=0.4)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^(n-x)$

Where (nCx) means combinatory and it's given by this formula:

$nCx=(n!)/((n-x)! x!)$

3) Part a

$P(X=13)=(22C13)(0.4)^(13) (1-0.4)^(22-13)=0.0336$

$P(X=17)=(22C17)(0.4)^(17) (1-0.4)^(22-17)=0.00035$

$P(X=13 U X=17) = 0.0336+0.00035=0.0340$

4) Part b

$P(X \geq 17)=P(X=17)+P(X=18)+P(X=19)+P(X=20)+P(X=21)+P(X=22)$

$P(X=17)=(22C17)(0.4)^(17) (1-0.4)^(22-17)=0.00035$

$P(X=18)=(22C18)(0.4)^(18) (1-0.4)^(22-18)=0.0000651$

$P(X=19)=(22C19)(0.4)^(19) (1-0.4)^(22-19)=0.00000914$

$P(X=20)=(22C20)(0.4)^(20) (1-0.4)^(22-20)=0.000000914$

$P(X=21)=(22C21)(0.4)^(21) (1-0.4)^(22-21)=5.81x10^(-8)$

$P(X=22)=(22C22)(0.4)^(22) (1-0.4)^(22-22)=1.76x10^(-9)$

$P(X \geq 17)=0.000427$

5) Part c

$P(X \leq 16)=1-P(X>16)=1-P(X\geq 17) = 1-[P(X=17)+P(X=18)+P(X=19)+P(X=20)+P(X=21)+P(X=22)]$

$P(X \geq 16)= 1-0.000427=0.9996$

6) Part d

The expected value is:

$E(X)= np = 22*0.4=8.8$

7) Part e

The variance is given by:

$Var(X) = np(1-p)= 22*0.4*(1-0,4)=5.28$

Find the sum of the first six terms of a geometric progression .

Answers

Question:

Find the sum of the first six terms of a geometric progression.

1,3,9,....

Answer:

$S_6 = 364$

Step-by-step explanation:

For a geometric progression, the sum of n terms is:

$S_n = (a(r^n - 1))/(r - 1)$

In the given sequence:

$a = 1$

$r = 3/1 =3$

$n = 6$

So:

$S_n = (a(r^n - 1))/(r - 1)$

$S_6 = (1 * (3^6 - 1))/(3 - 1)$

$S_6 = (3^6 - 1)/(2)$

$S_6 = (728)/(2)$

$S_6 = 364$

Oct 16, 10:30:54 AMA rocket is shot into the air. The function f (x) = -16x2 + 64x + 8 gives the
height of the rocket (in feet) as a function of the rockets horizontal distance from
where it was initially shot.
a. What was the initial height of the rocket when it was shot?
b. What is the maximum height the rocket reaches in the air?
a. The initial height of the rocket was
feet.
b. The maximum height the rocket reaches is
feet.