Mathematics, 24.01.2020 21:31 ithebrainliest

Suppose that the mean value of interpupillary distance for all adult males is 65 mm, and the population standard deviation is 5 mm.

a) find the probability that a male selected at random will have an interpupillary distance less that 67.5 mm. assume that the distribution of interpupillary distances is approximately normal.

b) if a sample of 22 adult males is selected, what is the probability that the sample average distance x is between 64.3 and 66.4 mm?

c) given that the sample size in part (b) is small, explain why we could still use the z-table to answer the question.

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Answer from: dozsyerra

a) P(X

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

P(z

b) P(64.3

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.

Answer from: Quest

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Answer from: Quest

a and c

step-by-step explanation:

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