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A control chart is developed to monitor the analysis of iron levels in human blood. The lines on the control chart were obtained

sergejj [24]

Answer:

Step-by-step explanation:

Hello!

The variable is X: iron level on human blood.

It has a mean of μ= 51.50 mg/dl and a standard deviation of σ= 3.50 mg/dl.

According to the control chart, the process should be shut down for troubleshooting when the analysis shows values X[bar]≥ 53.42 mg/dl and X[bar]≤ 49.58 mg/dl

Warnings are received at levels X[bar]≥52.78 mg/dl and X[bar]≤ 50.22 mg/dl

The system works between levels 49.58<X[bar]<53.42 and works without warnings between 50.22<X[bar]<52.78.

Using these parameters you have to analyze if the lists of sample means to see which ones are within the working values are wich ones are outside this interval.

Sample 1

Min= 50.15

Max= 51.99

Mean= 51.61

Without warning 50.22<X[bar]<52.78

Without Action interval 49.58<X[bar]<53.42

This sample's min value is below the lower limit of the warning interval but not low enough to reach action levels, the max value is within the working range.

Sample 2

Min= 50.32

Max= 52.56

Mean= 51.16

Without warning 50.22<X[bar]<52.78

Without Action interval 49.58<X[bar]<53.42

Both the max and min values of the sample are within the working range without warning.

Sample 3

Min= 50.25

Max= 53.12

Mean= 51.83

Without warning 50.22<X[bar]<52.78

Without Action interval 49.58<X[bar]<53.42

The max value of this sample is above the upper limit of the warning interval but does not surpass the upper bond of the troubleshoot interval. Min value is within working values.

Sample 4

Min= 50.05

Max= 53.01

Mean= 51.70

Without warning 50.22<X[bar]<52.78

Without Action interval 49.58<X[bar]<53.42

Both min and max values surpass the warning interval but do not reach action levels.

Sample 5

Min= 50.35

Max= 52.71

Mean= 51.37

Without warning 50.22<X[bar]<52.78

Without Action interval 49.58<X[bar]<53.42

Both min and max values are within working levels without warnings.

Considering that samples reaching warning levels should be shut down as a precaution, they are classified as:

Shutdown: Sample 1, 3 and 4

Do not shutdown: Sample 2 and 5.

I hope it helps!

8 0

2 years ago

Suppose a1=12−12,a2=23−13,a3=34−14,a4=45−15,a5=56−16. a) Find an explicit formula for an: . b) Determine whether the sequence is

lorasvet [3.4K]

I'm going to assume you meant to write fractions (because if $a_n$ are all non-negative integers, the series would clearly diverge), so that

$a_1=\dfrac12-\dfrac12$

$a_2=\dfrac23-\dfrac13$

$a_3=\dfrac34-\dfrac14$

and so on.

a. If the pattern continues as above, we would have the general term

$a_n=\dfrac n{n+1}-\dfrac1{n+1}=\dfrac{n-1}{n+1}$

b. Note that we can write $a_n$ as

$a_n=\dfrac{n-1}{n+1}=\dfrac{n+1-2}{n+1}=1-\dfrac2{n+1}$

The series diverges by comparison to the divergent series

$\displaystyle\sum_{n=1}^\infty\frac1n$

4 0

2 years ago

Implicit information requires the reader to combine details from the text with background knowledge to make a(n) . -guess -theor

klio [65]

The correct answer that would best complete the given statement above would be the third option: INFERENCE. Implicit information requires the reader to combine details from the text with background knowledge to make an inference. An inference is a conclusion that is based on evidence or reasoning.

3 0

2 years ago

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Estimate how many 12 inch rulers will be about the same length as a bike

navik [9.2K]

Answer:

So that means length of the bike is approx 5.7 rulers.

Step-by-step explanation:

the question says to estimate how many 12-inch rulers will be about the same length as a bike. In other words we have to measure the length of bike by in terms of ruler like 3 ruler length or 5 rular length etc.

Type of bike is not mentioned so i'm going to use bicyle.

From internet I found that approx lenght of a bike ( bicycle) is = 68 inches.

Given that 1 ruler = 12 inches

So number of rulers that can fit into 68 inches can be found by dividing 68 by 12

68/12 = 5.7

8 0

2 years ago

In USA Today (Sept. 5, 1996), the results of a survey involving the use of sleepwear while traveling were listed as follows: Mal

Wewaii [24]

Answers:

a) 0.018.

b) 0.614.

c) 0.166.

d) 0.479

Explanations:

1) Table

----------------------Male ----------- Female ---------- Total

Underwear ----- 0.220 ---------- 0.024 ---------- 0.244

Nightgown ----- 0.002 ---------- 0.180 ------------ 0.182

Nothing --------- 0.160 ----------- 0.018 ------------ 0.178

Pajamas --------- 0.102 ---------- 0.073 ----------- 0.175

T-shirt ------------ 0.046 --------- 0.088 ------------ 0.134

Other --------------0.084 --------- 0.003 ------------ 0.087

2) Check that it is a table of relative frequencies:

total = 0.244 + 0.182 + 0.178 + 0.175 + 0.134 + 0.087 = 1 ⇒ indeed this is a table of relative frequencies.

That means that each data is a probabiilty.

3) Soluitions:

(a) What is the probability that a traveler is a female who sleeps in the nothing?

relative frequency of females that use nothing = 0.018

(b) What is the probability that a traveler is male?

Probability that a traveler is a male = relative frequency of males = 0.220 + 0.002 + 0.160 + 0.102 + 0.046 + 0.084 = 0.614

(c) Assuming the traveler is male, what is the probability that he sleeps in pajamas?

It is the relative frequency of the males that sleep in pajamas divided by the relative frequency of males = 0.102 / 0.614 = 0.166

(d) What is the probability that a traveler is male if the traveler sleeps in pajamas or a T-shirt?

It is the relative frequency of males that sleep in pajamas plus that of the males that sleep in T-shirt, divided the total relative frequency of travelers that sleep in pajamas or T-shirt

i) relative frequency of males that sleeps in pajamas plus that of the males that sleeps in T-shirt = 0.102 + 0.046 = 0.148

ii) relative frequency of travelers that sleep in pajamas or T-shirt = 0.102 + 0.046 + 0.073 + 0.088 = 0.309

iii) quotient = 0.148 / 0.309 = 0.479

5 0

2 years ago

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