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guapka [62]
2 years ago
8

What is the estimate of 1374 x 6

Mathematics
1 answer:
timama [110]2 years ago
7 0
Hey it is 1374 x6 8244

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After dinner 2/3 of the corn bread is left suppose 4 friends want to share it equally. What fraction names how much of whole pan
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Each person would get 1/6 of the pan of corn bread that was left.  2/3 divided by 4= 1/6
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Automated Teller Machines (ATMs) provide all the following services except __.
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Most ATM machines at a bank will allow you to deposit money, transfer money and check your account balance.
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50 points decreased by 26% is how many points?
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At a Psychology final exam, the scores are normally distributed with a mean 73 points and a standard deviation of 10.6 points. T
USPshnik [31]

Answer:

The score that separates the lower 5% of the class from the rest of the class is 55.6.

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}

The Z-score measures how many standard deviations the measure is from the mean. 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, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question:

\mu = 73, \sigma = 10.6

Find the score that separates the lower 5% of the class from the rest of the class.

This score is the 5th percentile, which is X when Z has a pvalue of 0.05. So it is X when Z = -1.645.

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

-1.645 = \frac{X - 73}{10.6}

X - 73 = -1.645*10.6

X = 55.6

The score that separates the lower 5% of the class from the rest of the class is 55.6.

3 0
2 years ago
A consensus forecast is the average of a large number of individual analysts' forecasts. Suppose the individual forecasts for a
zaharov [31]

Answer:

a) P(X>3.5)=P(\frac{X-\mu}{\sigma}>\frac{3.5-\mu}{\sigma})=P(Z>\frac{3.5-5}{1.2})=P(z>-1.25)

And we can find this probability using the complement rule:

P(z>-1.25)=1-P(z

b) P(X

And we can find this probability uing the normal standard table:

P(z

c) P(3.5

And we can find this probability with this difference:

P(-1.25

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

P(-1.25

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

X \sim N(5,1.2)  

Where \mu=5 and \sigma=1.2

We are interested on this probability

P(X>3.5)

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>3.5)=P(\frac{X-\mu}{\sigma}>\frac{3.5-\mu}{\sigma})=P(Z>\frac{3.5-5}{1.2})=P(z>-1.25)

And we can find this probability using the complement rule:

P(z>-1.25)=1-P(z

Part b

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 uing the normal standard table:

P(z

Part c

P(3.5

And we can find this probability with this difference:

P(-1.25

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

P(-1.25

3 0
2 years ago
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