A 95% confidence interval is found to be (28, 32). The sample standard

Jumbo eggs in Australia, on average, are supposed to weigh 68g. Margot is in charge of a quality control test that involves weig

Andrej [43]

Answer:

0.9987 = 99.87% probability that the mean weight of 4 eggs in a package is less than 68.5g

Step-by-step explanation:

To solve this question, we use 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.

Mean weight of 67g and a sample standard deviation of 1g.

This means that $\mu = 67, \sigma = 1$

Sample of 4

This means that $n = 4, s = \frac{1}{\sqrt{4}} = 0.5$

What is the probability that the mean weight of 4 eggs in a package is less than 68.5g?

This is the pvalue of Z when X = 68.5. So

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

By the Central Limit Theorem

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

$Z = \frac{68.5 - 67}{0.5}$

$Z = 3$

$Z = 3$ has a pvalue of 0.9987

0.9987 = 99.87% probability that the mean weight of 4 eggs in a package is less than 68.5g

6 0

3 years ago

Read 2 more answers

HELP!! need answers for E and F!

BigorU [14]

Ok ok ok ok ok ok ok lol sorry don’t know

8 0

3 years ago

In a rhombus MPKN with an obtuse angle K the diagonals intersect each other at point E. The measure of one of the angles of a ∆P

allsm [11]

The rhombus has a couple of very interesting properties. The first is that the diagonals meet at 90o angles.

The second is that all the sides are congruent. That's actually the key to the problem (or one of them.

The third is that the diagonals are line segments that bisect the angles where the vertex of the angle is.

So just to make sure you understand what that last statement means <PKM = <NKM
K is the vertex of angle PKN

Now the really heavy duty stuff about your question.
Given
K is obtuse. Therefore <PKM can't be 16o because MKN would also have to be 16 degrees and together they don't add up to anything over 90o.
So the 16o angle is at <EPK

Remember that the diagonals meet at right angles. <PEK = 90o

Finally all triangles have 180o
< PKE = 180 - 16 - 90 = 74. So to review
<PKE = 74o
<EPK = 16o
<PEK = 90o That's one half the problem.

Moving on to triangle PMN
By the properties of parallel lines and a rhombus and isosceles triangles, that since PKN is bisected (rhombus property) Then PKN = 2* PKM =2*74 = 148o
The angle opposite <PKN is equal to <PKN so <PMN = <PKN
Since PKN = 148 then PMN = 148
Since KPN = 16o then PMN = 16o
Since triangle <PMN is isosceles <PNM = 16o

Summing up
PMN = 148o
MPN = 16o
MNP = 16o

That's both triangles solved. This is a really nice little problem. If you google properties of a rhombus, you will find all the properties I have used proven.

5 0

4 years ago

Read 2 more answers

Part of a regression output is provided below. Some of the information has been omitted.

quester [9]

Answer:

301.189

Step-by-step explanation:

Given the table :

Source of Variation - - SS - - - df - - - MS - - - - F

Regression - - - - - -3177.17 - - -2 - - 1588.6

Residual - - - - - - - - ______ --17 - - -17.717

Total - - - - - - - - - - 3478.36 - - 19

Calculate the SSR, Sum of Square residual

The Sum of Square RESIDUAL (SSR), Mean Square Residual (MSR) and Degree of Freedom RESIDUAL (DFR) are related by the formular :

MSR = SSR / DFR

Hence,

SSR = MSR × DFR

Fr the table ;

MSR = 17.717 ; DFR = 17

SSR = (17.717 × 17)

SSR = 301.189

5 0

3 years ago

If 8 g of a radioactive substance are present initially and 2 yr later only 4 g remain, how much of the substance will be presen

My name is Ann [436]

Answer:

none

Step-by-step explanation:

8-4=4

if every 2 years, 4 gallons disappear

then every 1 year, 2 gallons disappear

meaning that by 5 years there will be no substance left.

(5*2=10)

7 0

3 years ago