What is the RANGE and the IQR for 17,18,22,17,16,21,19,and16

The average of 100 measurements is 23 and the average of 50 additional measurements is 27.

Elden [556K]

Total value of 100 measurements
100 * 23
2300

Total value of additional 50 measurements
50 * 27
1350

Total value of 150 measurements
2300 + 1350
3650

Average of 150 measurements
3650 / 150
24.3

25 > 24.3

Answer is B

8 0

3 years ago

Make ”a” the subject of the formula 9a-8=2(a-b)

beks73 [17]

Answer:

Step-by-step explanation:

9a - 8 = 2a - 2b

7a - 8 = -2b

7a = 8 - 2b

a = (8 - 2b)/7

3 0

3 years ago

It is known that the life of a particular auto transmission follows a normal distribution with mean 72,000 miles and standard de

scoray [572]

Answer:

a) $P(X$

$P(z$

b) $P(X>65000)=P(\frac{X-\mu}{\sigma}>\frac{65000-\mu}{\sigma})=P(Z>\frac{65000-72000}{12000})=P(z>-0.583)$

$P(z>-0.583)=1-P(Z$

c) $P(X>100000)=P(\frac{X-\mu}{\sigma}>\frac{100000-\mu}{\sigma})=P(Z>\frac{100000-72000}{12000})=P(z>2.33)$

$P(z>2.33)=1-P(Z$

Sicne this probability just represent 1% of the data we can consider this value as unusual.

d) $z=1.28$

And if we solve for a we got

$a=72000 +1.28*12000=87360$

So the value of height that separates the bottom 90% of data from the top 10% is 87360.

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 life of a particular auto transmission of a population, and for this case we know the distribution for X is given by:

$X \sim N(72000,12000)$

Where $\mu=72000$ and $\sigma=12000$

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 using excel or the normal standard table and we got:

$P(z$

Part b

$P(X>65000)$

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:

$P(X>65000)=P(\frac{X-\mu}{\sigma}>\frac{65000-\mu}{\sigma})=P(Z>\frac{65000-72000}{12000})=P(z>-0.583)$

And we can find this probability using the complement rule and excel or the normal standard table and we got:

$P(z>-0.583)=1-P(Z$

Part c

$P(X>100000)$

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:

$P(X>100000)=P(\frac{X-\mu}{\sigma}>\frac{100000-\mu}{\sigma})=P(Z>\frac{100000-72000}{12000})=P(z>2.33)$

And we can find this probability using the complement rule and excel or the normal standard table and we got:

$P(z>2.33)=1-P(Z$

Sicne this probability just represent 1% of the data we can consider this value as unusual.

Part d

For this part we want to find a value a, such that we satisfy this condition:

$P(X>a)=0.1$ (a)

$P(X$ (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.9 of the area on the left and 0.1 of the area on the right it's z=1.28. On this case P(Z<1.28)=0.9 and P(z>1.28)=0.1

If we use condition (b) from previous we have this:

$P(X$

$P(z$

But we know which value of z satisfy the previous equation so then we can do this:

$z=1.28$

And if we solve for a we got

$a=72000 +1.28*12000=87360$

So the value of height that separates the bottom 90% of data from the top 10% is 87360.

5 0

3 years ago

A scale measures weight to the nearest 0.1 pounds. Which is the most appropriate way to report weight using this scale?

Pani-rosa [81]

The question is kind of vague in my opinion.

the best measurement to report with the scale in my opinion would be tens of pounds. my reasoning is because the decimal place would still prove importance with numbers under one hundred. above one hundred pounds and the decimal place becomes kind of pointless.

I hope I answered your question. though i'm sorry if i didn't

8 0

3 years ago

A manufacturing company regularly conducts quality control checks at specified periods on the products it manufactures. Historic

Rina8888 [55]

Answer:

The probability that none of the LED light bulbs are defective is 0.7374.

Step-by-step explanation:

The complete question is:

What is the probability that none of the LED light bulbs are defective?

Solution:

Let the random variable X represent the number of defective LED light bulbs.

The probability of a LED light bulb being defective is, P (X) = p = 0.03.

A random sample of n = 10 LED light bulbs is selected.

The event of a specific LED light bulb being defective is independent of the other bulbs.

The random variable X thus follows a Binomial distribution with parameters n = 10 and p = 0.03.

The probability mass function of X is:

$P(X=x)={10\choose x}(0.03)^{x}(1-0.03)^{10-x};\ x=0,1,2,3...$

Compute the probability that none of the LED light bulbs are defective as follows:

$P(X=0)={10\choose 0}(0.03)^{0}(1-0.03)^{10-0}$

$=1\times 1\times 0.737424\\=0.737424\\\approx 0.7374$

Thus, the probability that none of the LED light bulbs are defective is 0.7374.

6 0

3 years ago