# As the saying goes, “You can't please everyone.” Studies have shown that in a largepopulation approximately 4.5% of the population will be displeased, regardless of thesituation. If a random sample of 25 people are selected from such a population, what is theprobability that at least two will be displeased?A) 0.045B) 0.311C) 0.373D) 0.627E) 0.689

The probability that at least two people will be displeased in a random sample of 25 people is approximately 0.202.

### What is probability?

It is the chance of an event to occur from a total number of outcomes.

The formula for probability is given as:

Probability = Number of required events / Total number of outcomes.

Example:

The probability of getting a head in tossing a coin.

P(H) = 1/2

We have,

This problem can be solved using the binomialdistribution since we have a fixed number of trials (selecting 25 people) and each trial has two possible outcomes (displeased or not displeased).

Let p be the probability of an individual being displeased, which is given as 0.045 (or 4.5% as a decimal).

Then, the probability of an individual not being displeased is:

1 - p = 0.955.

Let X be the number of displeasedpeople in a random sample of 25.

We want to find the probability that at least two people are displeased, which can be expressed as:

P(X ≥ 2) = 1 - P(X < 2)

To calculate P(X < 2), we can use the binomial distribution formula:

where n is the samplesize (25), k is the number of displeasedpeople, and (n choose k) is the binomial coefficient which represents the number of ways to choose k items from a set of n items.

For k = 0, we have:

≈ 0.378

For k = 1, we have:

≈ 0.42

Therefore,

P(X < 2) = P(X = 0) + P(X = 1) ≈ 0.798.

Finally, we can calculate,

P(X ≥ 2) = 1 - P(X < 2)

= 1 - 0.798

= 0.202.

Thus,

The probability that at least two people will be displeased in a random sample of 25 people is approximately 0.202.

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Step-by-step explanation:

Let X = the number of people that are displeased in a random sample of 25 people selected from a population of which 4.5% will be displeased regardless of the situation. Then X is a binomial random variable with n = 25 and p = 0.045.

P(X ≥ 2) = 1 – P(X ≤ 1) = 1 – binomcdf(n: 25, p: 0.045, x-value: 1) = 0.311.

P(X ≥ 2) = 1 – [P(X = 0) + P(X = 1)] = 1 – 0C25(0.045)0(1 – 0.045)25 – 25C1(0.045)1(1 – 0.045)24 = 0.311.

## Related Questions

Through the points (1/2 , 3/4) and (-1/3 , 5/4).Question 2
Find, if possible, the slope of the line through the points (2 , 5) and (-4 , 5)

0

Step-by-step explanation:

m=(y2-y1)/(x2-x1)

m=(5-5)/(-4-2)

m=0/-6

m=0

Step-by-step explanation:

Every perpendicular corner of a square is 90 degrees.

Lines are cut half diagonally.

So

Use a tree diagram or systematic list to determine the number of ways a nickel, a dime, and a quarter can be flipped to get exactly one tail. Remember outcomes can be described with strings of T's and H's.

The Tree diagram is shown below.

### What is Tree Diagram?

There are twoparts to a probability tree diagram. These are given as follows:

• Node - it is used to represent an event.
• Branch - it is used to connect two events. The probability of occurrence of an event is written on a branch.

Given:

The tree diagram for Nickel, a dime, and a quarter is shown below

Nickle                   Dime               Quarter                 Outcome

/                             \       Tail----------------> HHT

/

\

\

\     Tail----------------> HTT

/                      \   Tail----------------> THT

/

\                      /

Tail

\

Tail----------------> TTT

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is there any options I can chose from

According to the Insurance Institute of America, a family of four spends between $400 and$3,800 per year on all types of insurance. Suppose the money spent is uniformly distributed between these amounts.a. What is the mean amount spent on insurance?
b. What is the standard deviation of the amount spent?
c. If we select a family at random, what is the probability they spend less than $2,000 per year on insurance per year? d. What is the probability a family spends more than$3,000 per year?

b. 981.5

c. 0.471

d. 0.235

Step-by-step explanation:

Given: According to the Insurance Institute of America, a family of four spends between $400 and$3,800 per year on all types of insurance.

If the money spent is uniformly distributed between these amounts.

Let A=$400 and B=$3,800

a. The mean amount spent on insurance =

b. The standard deviation of the amount spent=

c. To calculate, the probability they spend less than $2,000 per year on insurance per year ,it means the amount is between 400 and 2000 i.e Amount=$2000-$400=$1600

Then P(400<insurance amount<2000)

d. Ia a family spends more than $3000 then the amount is between$3000 and $3800, i.e. Amount=$3800-$3000=$800

Now,

P(3000<insurance amount<3800)=

a. Because of the uniform distribution, the mean = the middle of both numbers, or 1/2(3800 + 400) = \$2100.
b. To find the standard deviation, we take the variance ((b-a)/√12) and plug in the values:
(3800 - 400)/√12
3400/√12
~981.5
c. Again, because of uniform distribution, we just have to find the percentage that all numbers less than 2000 have in the range of 400 to 3800. To make things simple, I'm going to subtract 400 from both 3800 and 2000 so the data cuts off at 0. The equation is:
1600 = 3400 * x
16/34 = x (simplified from 1600/3400)
8/17 = x
d. same thing with c, but slightly different:
3800 - 400 = 3400, 3000 - 400 = 2600
3400 - 2600 = 800
800 = 3400 * x
8/34 = x
4/17 = x

Suppose Jesse won five bags of eight goldfish use math you know to represent the problem and find the number of goldfish Jesse won

40

Step-by-step explanation:

5(8) = x

use 5 as bags and use 8 as number of goldfish

then times the 2 number

What are the answers to 7 and 8?