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ankoles [38]
3 years ago
15

Average of 6 numbers is 4, if the average of 2 numbers is 2 what is the average of other 4

Mathematics
1 answer:
RideAnS [48]3 years ago
4 0

Answer:

5

Step-by-step explanation:

Let's just pretend the 6 numbers are: a,b,c,d,e,f.

Then (a+b+c+d+e+f)/6 =4 (given by first statement.)

Let's pretend the average of 2 numbers is 2 means they are talking about the a and b.

So (a+b)/2=2.

Now we want to find (c+d+e+f)/4 as requested by their question.

So first step multiply both sides of our first equation by 6 giving us:

a+b+c+d+e+f=24

So the second step multiply both sides of our second equation by 2 giving us:

a+b=4.

Now if a+b+c+d+e+f=24 and a+b=4, then c+d+e+f=20 since 20+4=24.

So the average of the four numbers c,d,e, and f is 20/4=5.

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Answer:

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Step-by-step explanation:

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natta225 [31]

Answer:

a) 99% of the sample means will fall between 0.933 and 0.941.

b) By the Central Limit Theorem, approximately normal, with mean 0.937 and standard deviation 0.0015.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions 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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

(a) If the true mean is 0.9370 with a standard deviation of 0.0090 within what interval will 99% of the sample means fail?

Samples of 34 means that n = 34

We have that \mu = 0.937, \sigma = 0.009

By the Central Limit Theorem, s = \frac{0.009}{\sqrt{34}} = 0.0015

Within what interval will 99% of the sample means fail?

Between the (100-99)/2 = 0.5th percentile and the (100+99)/2 = 99.5th percentile.

0.5th percentile:

X when Z has a pvalue of 0.005. So X when Z = -2.575.

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

By the Central Limit Theorem

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

-2.575 = \frac{X - 0.937}{0.0015}

X - 0.937 = -2.575*0.0015

X = 0.933

99.5th percentile:

X when Z has a pvalue of 0.995. So X when Z = 2.575.

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

2.575 = \frac{X - 0.937}{0.0015}

X - 0.937 = 2.575*0.0015

X = 0.941

99% of the sample means will fall between 0.933 and 0.941.

(b) If the true mean 0.9370 with a standard deviation of 0.0090, what is the sampling distribution of ¯X?

By the Central Limit Theorem, approximately normal, with mean 0.937 and standard deviation 0.0015.

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Marshall and Peter went to lunch at a cafe. They ordered a spinach salad for $4.15, a tuna sandwich for $6.35, and 2 glasses of
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Answer:

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Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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