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zvonat [6]
3 years ago
15

For a standard normal distribution, find the approximate value of P (z greater-than-or-equal-to negative 1. 25). Use the portion

of the standard normal table below to help answer the question. Z Probability 0. 00 0. 5000 0. 25 0. 5987 1. 00 0. 8413 1. 25 0. 8944 1. 50 0. 9332 1. 75 0. 9599 11% 39% 61% 89%.
Mathematics
1 answer:
alekssr [168]3 years ago
3 0

Answer:

89%

Step-by-step explanation:

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374 I think 15 x 22 = 330 + 44 =374

Step-by-step explanation:

I think 15 x 22 = 330 + 44 =374

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Solve the following proportion (solve p) p/6=7/8
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Given the following information, calculate the
Anna11 [10]

Answer:

150

Step-by-step explanation:

The ratio of ZY to SW is 16:24, meaning it is 2:3, so all of the sides of RST are 1.5 times as long as RZX.

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6 0
2 years ago
Assume that the heights of men are normally distributed with a mean of 70.270.2 inches and a standard deviation of 2.12.1 inches
Wittaler [7]

Answer:

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

X \sim N(70.2,2.12)  

Where \mu=70.2 and \sigma=2.12

Since the disgtribution for X is normal then the distribution for the sample mean \bar X is given by:

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

We can find the probability of interest we got:

P(\bar X >71.2)=P(Z>\frac{71.2-70.2}{\frac{2.12}{\sqrt{36}}}=2.83)

And using the complement rule and a calculator, excel or the normal standard table we got this:

P(Z>2.83)=1-P(Z

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 heights of a population, and for this case we know the distribution for X is given by:

X \sim N(70.2,2.12)  

Where \mu=70.2 and \sigma=2.12

Since the disgtribution for X is normal then the distribution for the sample mean \bar X is given by:

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

We can find the probability of interest we got:

P(\bar X >71.2)=P(Z>\frac{71.2-70.2}{\frac{2.12}{\sqrt{36}}}=2.83)

And using the complement rule and a calculator, excel or the normal standard table we got this:

P(Z>2.83)=1-P(Z

5 0
3 years ago
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