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aliya0001 [1]
3 years ago
7

Choose all the numbers that are multiples of 7. 7 28 34 49 21 54 63

Mathematics
1 answer:
Alecsey [184]3 years ago
5 0
The numbers are 7, 28, 49, 21, 63
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Suppose that the mean value of interpupillary distance for all adult males is 65 mm, and the population standard deviation is 5
Ksenya-84 [330]

Answer:

a) P(X

And we can find this probability using the z table or excel:

P(z

b) P(64.3

P(-0.657

P(-0.657

c) And since the distribution of X is normal we can conclude that the distribution for the sample mean is also normal and for this reason we can use the z table no matter if the sample size is small.

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".  

Part a

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

X \sim N(65,5)  

Where \mu=65 and \sigma=5

We are interested on this probability

P(X

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}

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

P(X

And we can find this probability using the z table or excel:

P(z

Part b

For this case the distirbution of the sample mean is also normal since the distribution for the random variable X is normal and is given by:

\bar X \sim N(\mu, \sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}= \frac{5}{\sqrt{22}}=1.066)

And for this case we want this probability:

P(64.3

And we can find this probability on this way:

P(-0.657

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.  

P(-0.657

Part c

For this case since the sample mean is defined as:

\bar x = \frac{\sum_{i=1}^n X_i}{n}

If we find the expected value for this variable we got:

E(\bar X)= \frac{1}{n} \sum_{i=1}^n X_i =\frac{n\mu}{n}=\mu

And for the variance we have:

Var(\bar X) =\frac{1}{n^2} \sum_{i=1}^n Var(Xi) =\frac{n \sigma^2}{n^2}= \frac{\sigma^2}{n}

And for this reason the deviation is \frac{\sigma}{\sqrt{n}}

And since the distribution of X is normal we can conclude that the distribution for the sample mean is also normal and for this reason we can use the z table no matter if the sample size is small.

4 0
3 years ago
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