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Mice21 [21]
3 years ago
14

How many fifths are the in 60/100

Mathematics
1 answer:
tankabanditka [31]3 years ago
8 0
In order to calculate that, we need to divide it by 1/5 as follows:
60/100 / 1/5

We can re-write it as: 60/100 * 5 = 6/10 * 5 = 30/10 = 3

So, there are "3" 1/5ths in 60/100

Hope this helps!
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Given a population with a mean of muμequals=100100 and a variance of sigma squaredσ2equals=3636​, the central limit theorem appl
lakkis [162]

Answer:

a) \bar X \sim N(100,\frac{6}{\sqrt{25}}=1.2)

\mu_{\bar X}=100 \sigma^2_{\bar X}=1

b) P(\bar X >101)=1-P(\bar X

c) P(\bar X

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 Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations 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".  

The complement rule is a theorem that provides a connection between the probability of an event and the probability of the complement of the event. Lat A the event of interest and A' the complement. The rule is defined by: P(A)+P(A') =1

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

X \sim N(\mu=100,\sigma=6)  

And let \bar X represent the sample mean, by the central limit theorem, the distribution for the sample mean is given by:  

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

a. What are the mean and variance of the sampling distribution for the sample​ means?

\bar X \sim N(100,\frac{6}{\sqrt{25}}=1.2)

\mu_{\bar X}=100 \sigma^2_{\bar X}=1.2^2=1.44

b. What is the probability that x overbarxgreater than>101

First we can to find the z score for the value of 101. And in order to do this we need to apply the formula for the z score given by:  

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

If we apply this formula to our probability we got this:  

z=\frac{101-100}{\frac{6}{\sqrt{25}}}=0.833  

And we want to find this probability:

P(\bar X >101)=1-P(\bar X

On this last step we use the complement rule.  

c. What is the probability that x bar 98less than

First we can to find the z score for the value of 98.

z=\frac{98-100}{\frac{6}{\sqrt{25}}}=-1.67  

And we want to find this probability:

P(\bar X

5 0
3 years ago
A widget company produces 25 widgets a day,5 of which are defective.Find the probability of selecting 5 widgets from the 25 prod
MariettaO [177]

Answer:

0.292

Step-by-step explanation:

Combinations can be used to solve the following problem.

We are choosing 5 widgets from 25 = ²⁵C₅

We want to select zero widgets from defective widgets = ⁵C₀

From the 20 non-defective widgets we want to select 5 = ²⁰C₅

So the probability is:

P = ( ⁵C₀ * ²⁰C₅) / ²⁰C₅

P = (1 * 15504) / 53130

= 15504/ 53130

=0.292 ..

7 0
3 years ago
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