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Calculate the UCL and LCL for a mean chart. Assume the standard deviation is unknown. Round to 1 decimal.

Alexeev081 [22]

Answer and explanation:

Please find answer and explanation attached

6 0

3 years ago

Almost all medical schools in the United States require students to take the Medical College Admission Test (MCAT). To estimate

Leviafan [203]

Answer:

Probability of having student's score between 505 and 515 is 0.36

Given that z-scores are rounded to two decimals using Standard Normal Distribution Table

Step-by-step explanation:

As we know from normal distribution: z(x) = (x - Mu)/SD

where x = targeted value; Mu = Mean of Normal Distribution; SD = Standard Deviation of Normal Distribution

Therefore using given data: Mu (Mean) = 510, SD = 10.4 we have z(x) by using z(x) = (x - Mu)/SD as under:

In our case, we have x = 505 & 515

Approach 1 using Standard Normal Distribution Table:

z for x=505: z(505) = (505-510)/10.4 gives us z(505) = -0.48

z for x=515: z(515) = (515-510)/10.4 gives us z(515) = 0.48

Afterwards using Normal Distribution Tables and rounding the values to two decimals we find the probabilities as under:

P(505) using z(505) = 0.32

Similarly we have:

P(515) using z(515) = 0.68

Now we may find the probability of student's score between 505 and 515 using:

P(505 < x < 515) = P(515)-P(505) = 0.68 - 0.32 = 0.36

PS: The standard normal distribution table is being attached for reference.

Approach 2 using Excel or Google Sheets:

P(x) = norm.dist(x,Mean,SD,Commutative)

P(505) = norm.dist(505,510,10.4,1)

P(515) = norm.dist(515,510,10.4,1)

Probability of student's score between 505 and 515= P(515) - P(505) = 0.36

Download pdf

6 0

3 years ago

Please help with this question.

Nookie1986 [14]

Answer:

The answer is B

Hope this helped :)

7 0

4 years ago

Write an equation of the line containing the given point and perpendicular to the given line:

bazaltina [42]

Answer:

Step-by-step explanation:

The equation of a straight line can be represented in the slope intercept form as

y = mx + c

Where

m = slope = (change in the value of y in the y axis) / (change in the value of x in the x axis)

The equation of the given line is

2x+9y=5

9y = - 2x + 5

y = -2x/9 + 5/9

Comparing with the slope intercept form, slope = -2/9

If the line passing through the given point is perpendicular to the given line, it means its slope is the negative reciprocal of the slope of the given line.

Therefore, the slope of the line passing through (4,-9) is 9/2

To determine the intercept, we would substitute m = 9/2, x = 4 and y = -9 into y = mx + c. It becomes

- 9 = 9/2×4 + c = 18 + c

c = - 9 - 18 = - 27

The equation becomes

y = 9x/2 - 27

4 0

4 years ago

Josie recorded the average monthly temperatures for two cities in the state where she lives. For City 1, what is the mean of the

Elena-2011 [213]

Answer:

$\bar X_1 =\frac{30+ 38+ 66+ 78+ 47+ 75+ 35+ 45+ 56+ 29+ 49+ 37}{12}=48.75$

The mean average monthly temperature in City 1 is 48.75°F.

$MAD = \frac{\sum_{i=1}^n |X_i -\bar X|}{n} = \frac{160.5}{12}= 13.375$

The mean absolute deviation for the average monthly temperature in City 1 is 13.375°F

Step-by-step explanation:

For this case we have the following dataset given:

Average Monthly Temperatures for City 1 (° F)

30, 38, 66, 78, 47, 75, 35, 45, 56, 29, 49, 37

Average Monthly Temperatures for City 2 (° F)

15, 23, 51, 63, 32, 60, 20, 30, 41, 14, 34, 22

For this case the sample mean can be calculated with the following formula:

$\bar X =\frac{\sum_{i=1}^n X_i}{n}$

And replacing we got:

$\bar X_1 =\frac{30+ 38+ 66+ 78+ 47+ 75+ 35+ 45+ 56+ 29+ 49+ 37}{12}=48.75$

The mean average monthly temperature in City 1 is 48.75°F.

And now we can calculate the following values:

$|30-48.75| =18.75$

$|38-48.75| =10.75$

$|66-48.75| =17.25$

$|78-48.75| =29.25$

$|47-48.75| =1.75$

$|75-48.75| =26.25$

$|35-48.75| =13.75$

$|45-48.75| =3.75$

$|56-48.75| =7.25$

$|29-48.75| =19.75$

$|49-48.75| =0.25$

$|37-48.75| =11.75$

And the mean absolute deviation is given by:

$MAD = \frac{\sum_{i=1}^n |X_i -\bar X|}{n} = \frac{160.5}{12}= 13.375$

The mean absolute deviation for the average monthly temperature in City 1 is 13.375°F

8 0

4 years ago