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Vanyuwa [196]
4 years ago
10

1. Data are from a normal distribution and the mean is 20 with a standard deviation of 2. a. What % of observations fall between

18 and 22
Mathematics
1 answer:
tangare [24]4 years ago
7 0

Answer:

P(18

And we can find this probability with the normal standard distribution and we got:

P(-1

Step-by-step explanation:

Let X the random variable that represent the variable of a population, and for this case we know the distribution for X is given by:

X \sim N(20,2)  

Where \mu=20 and \sigma=2

We are interested on this probability

P(18

And the best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{x-\mu}{\sigma}

Using this formula we got:

P(18

And we can find this probability with the normal standard distribution and we got:

P(-1

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Answer:

900

Step-by-step explanation:

2pi = 360

1pi = 180

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3 years ago
The volume of B is 160 millilitres. Work out the volume of A.
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Answer:32

Step-by-step explanation: The volume of A is 32:)

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3 years ago
Past records indicate that the probability of online retail orders
Tcecarenko [31]

Answer:

a) Mean = 1.6, standard deviation = 1.21

b) 18.87% probability that zero online retail orders will turn out to be fraudulent.

c) 32.82% probability that one online retail order will turn out to be fraudulent.

d) 48.31% probability that two or more online retail orders will turn out to be fraudulent.

Step-by-step explanation:

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} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

The mean of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

In this problem, we have that:

p = 0.08, n = 20

a. What are the mean and standard deviation of the number of online retail orders that turn out to be fraudulent?

Mean

E(X) = np = 20*0.08 = 1.6

Standard deviation

\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{20*0.08*0.92} = 1.21

b. What is the probability that zero online retail orders will turn out to be fraudulent?

This is P(X = 0).

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

P(X = 0) = C_{20,0}.(0.08)^{0}.(0.92)^{20} = 0.1887

18.87% probability that zero online retail orders will turn out to be fraudulent.

c. What is the probability that one online retail order will turn out to be fraudulent?

This is P(X = 1).

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

P(X = 1) = C_{20,1}.(0.08)^{1}.(0.92)^{19} = 0.3282

32.82% probability that one online retail order will turn out to be fraudulent.

d. What is the probability that two or more online retail orders will turn out to be fraudulent?

Either one or less is fraudulent, or two or more are. The sum of the probabilities of these events is decimal 1. So

P(X \leq 1) + P(X \geq 2) = 1

We want P(X \geq 2)

So

P(X \geq 2) = 1 - P(X \leq 1)

In which

P(X \leq 1) = P(X = 0) + P(X = 1)

From itens b and c

P(X \leq 1) = 0.1887 + 0.3282 = 0.5169

P(X \geq 2) = 1 - P(X \leq 1) = 1 - 0.5169 = 0.4831

48.31% probability that two or more online retail orders will turn out to be fraudulent.

4 0
4 years ago
3. Melissa is making a piece for a stained glass window that has the shape and measures shown below. Point Cis the center of the
AlekseyPX

Answer:

184.26 i dont exacaly know but i think it is just let me know if im right or wrong

Step-by-step explanation:

7 0
3 years ago
When Sharon began shopping this morning, she had $40.00. She purchased five paperback books and had lunch. The books were all th
Tpy6a [65]

40-7 = 33

33-3.25 = 29.75

29.75/5 = 5.95 each

8 0
3 years ago
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