Please help this question

A GMC dealer believes that demand for 2005 Envoys will be normally distributed with a mean of 200 and standard deviation of 30.

irina1246 [14]

Answer:

The correct answer to the following question will be "280 units".

Step-by-step explanation:

The given values are:

Mean = 200

Standard deviation = 30

Let's ask that the supplier needs to reach 99.9% of the standard of operation. To reach this quality of operation, she has to preserve an inventory of

Now,

⇒ $Mean + 3\times sigma$

On putting the values, we get

⇒ $200+3\times 30$

⇒ $200+90$

⇒ $290 \ units$

290 units would then accomplish a level of service of 99.9%.

Half unpurchased units can indeed be purchased at $30000 i.e. no failure. If he requests 300, there will be losses on 5 products that are unsold.

Hence, ordering 280 units seems to be preferable

3 0

3 years ago

0.6 : 3.3 : 3/2 reduce this<br>the : means its a ratio <br>REDUCE THE RATIO!!!

Sati [7]

3:5
1:1
1.5 or 1:5 not sure

3 0

3 years ago

I NEED HELP ON 7, 9, AND 10! PLEASE

Black_prince [1.1K]

✧･ﾟ: *✧･ﾟ:* Hello ✧･ﾟ: *✧･ﾟ:*

༶•┈┈⛧┈♛ ♛┈⛧┈┈•༶

I will give you a explanation about how I got the answer!

Plus, I'm am so sorry if I answer this so long...

Hope It Helped! :)

4 0

3 years ago

Dr. E. Quation says the Infant Mortality Rate of Korea shown by the expression; 7(m + 3) - 2. Is equal to his criminal success r

GarryVolchara [31]

Answer:

The infant mortality rate in Korea based on equality provided is:

1.8

Step-by-step explanation:

Since the text mentions that the infant mortality rate in Korea is equal to the criminal success rate, the two expressions must equalize and clear the variable m, which is the rate in each of the expressions:

7 (m + 3) - 2 = 8m + 17.2
7m + 21 - 2 = 8m + 17.2
7m + 19 = 8m + 17.2
19 - 17.2 = 8m - 7m
1.8 = m

As you can see, once the equality of the expressions is solved, <u>a rate of 1.8</u> is obtained.

6 0

3 years ago

A sample of size 6 will be drawn from a normal population with mean 61 and standard deviation 14. Use the TI-84 Plus calculator.

AVprozaik [17]

Answer:

$X \sim N(61,14)$

Where $\mu=61$ and $\sigma=14$

Since the distribution for X is normal then the distribution for the sample mean is also normal and given by:

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

$\mu_{\bar X} = 61$

$\sigma_{\bar X}= \frac{14}{\sqrt{6}}= 5.715$

So then is appropiate use the normal distribution to find the probabilities for $\bar X$

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". The letter $\phi(b)$ is used to denote the cumulative area for a b quantile on the normal standard distribution, or in other words: $\phi(b)=P(z$

Solution to the problem

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:

$X \sim N(61,14)$

Where $\mu=61$ and $\sigma=14$

Since the distribution for X is normal then the distribution for the sample mean $\bar X$ is also normal and given by:

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

$\mu_{\bar X} = 61$

$\sigma_{\bar X}= \frac{14}{\sqrt{6}}= 5.715$

So then is appropiate use the normal distribution to find the probabilities for $\bar X$

8 0

3 years ago