This is a practice test but I don’t understand stand this one pls help fast

On May 3, 1999, 59 tornadoes hit a certain state in the largest tornado outbreak ever recorded for that state. Sixteen of these

pishuonlain [190]

The question is incomplete, here is the complete question:

On May 3, 1999, 59 tornadoes hit a certain state in the largest tornado outbreak ever recorded for that state. Sixteen of these were classified as strong (F2 or F3) or violent (F4 or F5). Length of Path (miles) Intensity:

a. 7 F.

b. 9 F2.

c. 16 F3.

d. 11 F3.

e. 8 F2.

f. 9 F2.

g. 5 F2.

h. 16 F4.

i. 15 F3.

j. 38 F5.

k. 42 F4.

l. 3 F2.

m. 8 F2.

n. 1 F2.

o. 14 F3.

p. 23 F3.

Part (a) Make a box and whisker plot

Part (b) Identify outliners. Remove them and redraw the box and whisker diagram

Part (c) How does the removal of outliners effect box and whisker plot? How does is effect the median of the set?

Answer:

Given below with explanations

Step-by-step explanation:

(a)

The data is first arranged in ascending order:

1, 3, 5, 7, 8, 8, 9, 9, 11, 14, 15, 16, 16, 23, 38, 42

Total values = 16

Minimum value = 1

Maximum value = 42

Lower Quartile = 1/4 x 16 = 4th value = 7 (1, 3, 5, 7, 8, 8, 9, 9, 11, 14, 15, 16, 16, 23, 38, 42)

Upper Quartile = 3/4 x 16 = 12th value = 16 (1, 3, 5, 7, 8, 8, 9, 9, 11, 14, 15, 16, 16, 23, 38, 42)

Median = 1/2 x 16 = 8th value = 9 (1, 3, 5, 7, 8, 8, 9, 9, 11, 14, 15, 16, 16, 23, 38, 42)

The Box and whisker plot is attached in the picture below:

(b)

Outliners are 38 and 42

1, 3, 5, 7, 8, 8, 9, 9, 11, 14, 15, 16, 16, 23

Total values = 14

Minimum value = 1

Maximum value = 23

Lower Quartile = 1/4 x 14 = 3.5th value = 6 (1, 3, 5, 7, 8, 8, 9, 9, 11, 14, 15, 16, 16, 23)

(5+7)/2

Upper Quartile = 3/4 x 14 = 10.5th value = 14.5 (1, 3, 5, 7, 8, 8, 9, 9, 11, 14, 15, 16, 16, 23)

(14+15)/2

Median = 1/2 x 14 = 7th value = 9 (1, 3, 5, 7, 8, 8, 9, 9, 11, 14, 15, 16, 16, 23)

The Box and whisker plot is attached in the picture below:

(c)

The lower and upper quartile values have changed after removing the outliners but the median remains same in this case

7 0

3 years ago

Suppose that 100,000 men were screened for prostate cancer for the first time. Of these, 4,000 men had a positive result on the

Virty [35]

An actual two-by-two table is a tabular representation containing two rows and two columns.

The columns consist of the tested True positive for prostate cancer and tested True Negative for prostate cancer
The rows consist of the predicted positive screening and predicted negative values

Mathematically, the set-up of the two-by-two table for this data can be computed as:

Tested True Positive for cancer True Negative Total

Predicted Positive 800 3200 4000

Predicted Negative 100 95900 96000

Total 900 99100 100000

The prevalence rate of prostate cancer in this population is:

$\mathbf{ =\dfrac{900}{100000}}$

$\mathbf{ =\dfrac{9}{1000}}$

= 9 per thousand.

The calculation of the sensitivity of this screening is as follows:

$\mathbf{=\dfrac{TP}{TP+PN_1}}$

where;

TP = True positive for cancer
PN₁ = Predicted Negative for true positive cancer

∴

$\mathbf{=\dfrac{800}{800+100}}$

= 0.889

= 88.9%

The interpretation shows that 88.9% are correctly identified to be actual positive for prostate cancer.

The calculation of the specificity of this screening is as follows:

$\mathbf{=\dfrac{PN_2}{PN_2+TN}}$

where;

TN = True positive for cancer
PN₂ = Predicted Negative for true negative cancer

∴

$\mathbf{=\dfrac{95900}{95900+3200}}$

= 0.9677

= 96.77%

The interpretation shows that 96.7% of an actual negative is correctly identified as such.

The positive predicted value of the screening test is computed as:

$\mathbf{= \dfrac{TP}{TP + TN}}$

$\mathbf{= \dfrac{800}{800 + 3200}}$

= 0.2

= 20%

The interpretation of the positive predicted value of this screening shows that 20% that are subjected to the diagnosis of positive prostate cancer truly have the disease.

Learn more about tabular representation here:

brainly.com/question/8307968

6 0

2 years ago

The sales of a sandwich store increased approx. linearly from $52,000 to $116,000 during the first five yrs of business. Write a

dalvyx [7]

From x=1 to x=5
x=1 is starting time
x=time
y=mx+b
m=slope
b=yintercept

slope=(y2-y1)/(x2-x1)
for points (x1,y1) and (x2,y2)

(1,52000) and (5,116000)
slope=(116000-52000)/(5-1)=64000/4=16000
y=16000x+b
find b
(1,52000)
52000=16000(1)+b
52000=16000+b
minus 16000 from both sides
36000=b

the equation is
y=16000x+36000

at 12 years, x=12
y=16000(x)+36000
y=16000(12)+36000
y=192000+36000
y=228000

sales at year 12 is $228,000

8 0

3 years ago

What does the line 5x 2y = 8 look like?

Arturiano [62]

It's going to be a positive slope line so I would say from left to right top left to bottom right. Good luck!

6 0

3 years ago

Lacy draws a diamond from a standard deck of 52 cards. Without replacing the first card, she then proceeds to draw a second card

Viktor [21]

Answer:

a) i) No

ii) Probability of drawing a diamond and then spade without replacement is 0.0637.

b) i) Yes

ii) Probability of drawing a diamond and then spade with replacement is 0.0625

Step-by-step explanation:

Part a) Without replacement:

i) Are these events independent?

No, because the both events are dependent. because a card taken out of the deck is not replaced back before taking out the other card so the chances of event occur change.

ii) Determine the probability of drawing a diamond and then a spade without replacement.

Probability of drawing a diamond and then spade = (Probability of drawing a diamond) * (Probability of drawing a spade)

Probability of drawing a diamond :

Total cards in deck = 52

Diamonds = 13

Probability of drawing a diamond = 13/52

Probability of drawing a spade:

Total cards in deck after taking diamond = 52 -1 =51

Spades = 13

Probability of drawing a spade = 13/51

Probability of drawing a diamond and then spade = 13/52 * 13/51

= 169/2652

=0.0637

So, Probability of drawing a diamond and then spade without replacement is 0.0637

Part b) With replacements

i) Are these events independent?

Input Yes or No:

Yes, these events are independent because a card taken out of the deck is replaced back before taking out the other card so the chances of event occur doesn't change.

ii) Determine the probability of drawing a diamond and then a spade with replacement.

Probability of drawing a diamond and then spade = (Probability of drawing a diamond) * (Probability of drawing a spade)

Probability of drawing a diamond :

Total cards in deck = 52

Diamonds = 13

Probability of drawing a diamond = 13/52

Probability of drawing a spade:

Total cards in deck after replacing diamond = 52

Spades = 13

Probability of drawing a spade = 13/52

Probability of drawing a diamond and then spade = 13/52 * 13/52

= 169/2704

=0.0625

So, Probability of drawing a diamond and then spade with replacement is 0.0625

7 0

4 years ago