Write an equation for the quadratic function whose vertex is (âˆ’2,10) and passes through (1,28)

A random sample of size 100 was taken from a population. A 94% confidence interval to estimate the mean of the population was co

Firlakuza [10]

Answer:

Step-by-step explanation:

1) The z value was determined using a normal distribution table. From the normal distribution table, the corresponding z value for a 94% confidence interval is 1.88

The correct option is D

2) if 70% of all internet users experience e-mail fraud. It means that probability of success, p

p = 70/100 = 0.7

q = 1 - p = 1 - 0.7 = 0.3

n = number of selected users = 50

Mean, u = np = 50×0.7 = 35

Standard deviation, u = √npq = √50×0.7×0.3 = 3.24

x = number of internet users

The formula for normal distribution is expressed as

z = (x - u)/s

We want to determine the probability that no more than 25 were victims of e-mail fraud. It is expressed as

P(x lesser than or equal to 25)

The z value will be

z = (25- 35)/3.24 = - 10/3.24 = -3.09

Looking at the normal distribution table, the corresponding z score is 0.001

P(x lesser than or equal to 25) = 0.001

7 0

2 years ago

A batch of 445 containers for frozen orange juice contains 3 that are defective. Two are selected, at random, without replacemen

Elan Coil [88]

Answer:

When Two containers are selected

(a) Probability that the second one selected is defective given that the first one was defective = 0.00450

(b) Probability that both are defective = 0.0112461

(c) Probability that both are acceptable = 0.986

2. When Three containers are selected

(a) Probability that the third one selected is defective given that the first and second one selected were defective = 0.002.

(b) Probability that the third one selected is defective given that the first one selected was defective and the second one selected was okay = 0.00451

(c) Probability that all three are defective = 6.855 x $10^{-8}$ .

Step-by-step explanation:

We are given that a batch of 445 containers for frozen orange juice contains 3 defective ones i.e.

Total containers = 445

Defective ones = 3

Non - Defective ones = 442 { Acceptable ones}

Two containers are selected, at random, without replacement from the batch.

(a) Probability that the second one selected is defective given that the first one was defective is given by;

Since we had selected one defective so for selecting second the available

containers are 444 and available defective ones are 2 because once

chosen they are not replaced.

Hence, Probability that the second one selected is defective given that the first one was defective = $\frac{2}{444}$ = 0.00450

(b) Probability that both are defective = P(first being defective) +

P(Second being defective)

= $\frac{3}{445} + \frac{2}{444}$ = 0.0112461

(c) Probability that both are acceptable = P(First acceptable) + P(Second acceptable)

Since, total number of acceptable containers are 442 and total containers are 445.

So, Required Probability = $\frac{442}{445}*\frac{441}{444}$ = 0.986

Three containers are selected, at random, without replacement from the batch.

(a) Probability that the third one selected is defective given that the first and second one selected were defective is given by;

Since we had selected two defective containers so now for selecting third defective one, the available total containers are 443 and available defective container is 1 .

Therefore, Probability that the third one selected is defective given that the first and second one selected were defective = $\frac{1}{443}$ = 0.002.

(b) Probability that the third one selected is defective given that the first one selected was defective and the second one selected was okay is given by;

Since we had selected two containers so for selecting third container to be defective, the total containers available are 443 and available defective containers are 2 as one had been selected.

Hence, Required probability = $\frac{2}{443}$ = 0.00451 .