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yaroslaw [1]
2 years ago
13

Describe each of the following values as (A) a discrete random variable, (B) a continuous random variable, or (C) not a random v

ariable:
1. Exact weight of quarters now in circulation in the United States
2. Shoe sizes of humans
3. Political party affiliations of adults in the United States
A. 1.C
2.A
3.В
B. 1.B
2.A
3.С
C. 1.A
2.C
3.В
D. 1.A
2.В
3.С
Mathematics
1 answer:
Lyrx [107]2 years ago
3 0

Answer:

(1) B

(2) A

(3) C

Step-by-step explanation:

A random variable is a variable that denotes a set of all the possible outcomes of a random experiment. It is denotes by a single capital letter such as X or Y.

There are two types of random variables.

  • Discrete random variable: These type of random variable takes finite number of values, such as 0, 1, 2, 3, 4, ... For example, number of girl child in a neighborhood.
  • Continuous random variable: These type of random variables takes infinite number of possible values. For example, the height, weight.

(1)

Exact weight of quarters now in circulation in the United States.

The variable weight is a continuous variable.

Thus, the exact weight of quarters now in circulation in the United States is a continuous random variable.

(2)

Shoe sizes of humans.

The shoe size of a person are discrete and finite values.

Thus, the shoe sizes of humans are discrete random variables.

(3)

Political party affiliations of adults in the United States.

This variable is not a quantitative variable.

It is a qualitative variable.

Thus, the political party affiliations of adults in the United States is no random variable.

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The average life of a bread-making machine is 7 years, with a standard deviation of 1 year. Assuming that the lives of these mac
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Answer:

a) P(6.4

b) a=7 +1.036*0.333=7.345

So the value of bread-making machine that separates the bottom 85% of data from the top 15% is 7.345.

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 central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Let X the random variable life of a bread making machine. We know from the problem that the distribution for the random variable X is given by:

X\sim N(\mu =7,\sigma =1)

We take a sample of n=9 . That represent the sample size.

From the central limit theorem we know that the distribution for the sample mean \bar X is also normal and is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

\bar X \sim N(\mu=7, \frac{1}{\sqrt{9}})

Solution to the problem

Part a

(a) the probability that the mean life of a random sample  of 9 such machines falls between 6.4 and 7.2

In order to answer this question we can use the z score in order to find the probabilities, the formula given by:

z=\frac{\bar X- \mu}{\frac{\sigma}{\sqrt{n}}}

The standard error is given by this formula:

Se=\frac{\sigma}{\sqrt{n}}=\frac{1}{\sqrt{9}}=0.333

We want this probability:

P(6.4

Part b

b) The value of x to the right of which 15% of the  means computed from random samples of size 9 would fall.

For this part we want to find a value a, such that we satisfy this condition:

P(\bar X>a)=0.15   (a)

P(\bar X   (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.85 of the area on the left and 0.15 of the area on the right it's z=1.036. On this case P(Z<1.036)=0.85 and P(Z>1.036)=0.15

If we use condition (b) from previous we have this:

P(X  

P(z

But we know which value of z satisfy the previous equation so then we can do this:

z=1.036

And if we solve for a we got

a=7 +1.036*0.333=7.345

So the value of bread-making machine that separates the bottom 85% of data from the top 15% is 7.345.

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