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antoniya [11.8K]
3 years ago
6

Twenty percent of the students in a class of 100 are planning to go to graduate school. The standard deviation of this binomial

distribution is
Mathematics
1 answer:
Mademuasel [1]3 years ago
5 0

Answer: 4

Step-by-step explanation:

For any binomial variable X having parameters n (total number of trials ) and p (probability of getting success in each event), the standard deviation is given by :-

\sigma=\sqrt{np(1-p)}

As per given , we have

n= 100 , p= 20%=0.20

Then, the standard deviation of this binomial distribution is :

\sigma=\sqrt{100(0.20)(1-0.20)}\\\\=\sqrt{100(0.16)}=4

Hence, the  standard deviation of this binomial distribution is <u>4</u>.

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3 years ago
The mean annual income for people in a certain city is 37 thousand dollars, with a standard deviation of 28 thousand dollars. A
Aloiza [94]

Answer:

P( 31 < \bar X< 41)

And we can ue the z score formula given by:

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

And using this formula we got for the limits:

z = \frac{31-37}{\frac{28}{\sqrt{50}}}= -1.515

z = \frac{41-37}{\frac{28}{\sqrt{50}}}= 1.01

So we want to find this probability:

P(-1.515

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".  

Solution to the problem

Let X the random variable that represent the annual income of a population, and for this case we know the following info:

\mu=37 and \sigma=28  and we are omitting the zeros from the thousand to simplify calculations

We select a sample size of n=50>30.

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".

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

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

And we want to find this probability:

P( 31 < \bar X< 41)

And we can ue the z score formula given by:

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

And using this formula we got for the limits:

z = \frac{31-37}{\frac{28}{\sqrt{50}}}= -1.515

z = \frac{41-37}{\frac{28}{\sqrt{50}}}= 1.01

So we want to find this probability:

P(-1.515

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3 years ago
What does x equal? help please
Pavel [41]
It might be like this 107+31 equal 138
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