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3 years ago

If µ = 400 and s = 100 the probability of selecting at random a score less than or equal to 370 equals ____.

1 answer:

3 0

You'll need to calculate the applicable z-score:
(given value of x) - (mean value)
z = -----------------------------------------------
standard deviation

370-400 -30
Here, z = --------------- = ---------- = -0.3
100 100

Use either a table of z scores or your calculator to determine the area under the standard normal curve to the left of z = -0.3.

According to my TI-83 Plus calculator, this area is 0.382.

The probability of selecting at random a score less than or equal to 370 is 0.382.

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1 year ago

A sample of size 45 will be drawn from a population with mean 53 and standard deviation 11. Use the TI-84 Plus calculator. (a) I

sukhopar [10]

Answer:

a) We have the standard deviation and the mean, so it is appropriate to use the normal distribution to find probabilities for x.

b) There is a 15.97% probability that x will be between 54 and 55.

c) The 47th percentile of x is $X = 52.877$ .

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by

$Z = \frac{X - \mu}{\sigma}$

After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.

A sample of size 45 will be drawn from a population with mean 53 and standard deviation 11.

This means that $\mu = 53$ .

We have to find the standard deviation of the sample, that is:

$\sigma = \frac{11}{\sqrt{45}} = 1.64$

(a) Is it appropriate to use the normal distribution to find probabilities for x?

We have the standard deviation and the mean, so it is appropriate to use the normal distribution to find probabilities for x.

(b) Find the probability that x will be between 54 and 55.

This is the pvalue of the Z score when X = 55 subtracted by the pvalue of the Z score when X = 54.

X = 55

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{55 - 53}{1.64}$

$Z = 1.22$

$Z = 1.22$ has a pvalue of 0.88877.

X = 54

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{54 - 53}{1.64}$

$Z = 0.61$

$Z = 0.61$ has a pvalue of 0.72907.

So, there is a 0.88877 - 0.72907 = 0.1597 = 15.97% probability that x will be between 54 and 55.

(c) Find the 47th percentile of x. Round the answer to at least two decimal places.

This is the value of X when Z has a pvalue of 0.47;

This is between $Z = -0.07$ and $Z = -0.08$ . So we use $Z = -0.075$ .

$Z = \frac{X - \mu}{\sigma}$

$-0.075 = \frac{X - 53}{1.64}$

$X = 52.877$

The 47th percentile of x is $X = 52.877$ .

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3 years ago

Consider a population with population proportion p, and a sample from the population with sample proportion pˆ. Which of the fol

Deffense [45]

According to CollegeBoard.org:

<u>Answer</u>: "To estimate the probability of observing a value as extreme as pˆ given p."

<u>Explanation</u>: "The test statistic for a one-sample z-test is the distance, in units of standard deviations, between the statistic and the given parameter. From that distance, probabilities (a p-value) can be calculated and a claim can be assessed."

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2 years ago