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Blababa [14]
3 years ago
15

Potatoes: Suppose the weights of Farmer Carl's potatoes are normally distributed with a mean of 9 ounces and a standard deviatio

n of 1.2 ounces. Round your answers to 4 decimal places.
(a) If one potato is randomly selected, find the probability that it weighs less than 10 ounces.


(b) If one potato is randomly selected, find the probability that it weighs more than 12 ounces.


(c) If one potato is randomly selected, find the probability that it weighs between 10 and 12 ounces.

Mathematics
1 answer:
katrin2010 [14]3 years ago
5 0

Answer: a)  79.10 %           b) 7%                c)19.85 %

Step-by-step explanation:

a) Z = ( × - μ ) ÷ σ           Where  μ mean of population σ standard               deviation Z is the abscissa to give the area or probability we are looking for associated to the value 10 ounces ( × )

So:  Z = ( 10 - 9 ) ÷ 1.2  ⇒ Z = 1/1.2 ⇒   Z = 0.83  it has to be below this value

From Z tables we get  P [ Z ≤ 0.82 ] = 0.7910  0r  79.10 %

b) Following the same procedure: We look for

P [ Z > ( x - μ ) ÷   σ ]    ⇒   Z  =  ( 12 -10 ) ÷ 1.2  = 2.5

From Z table we get the area under the curve from the left tail up to the point Z < 2.5 ( 2.5 not included) but we were asked for the area out of that previous so 1- 0.9930 = 0.007 is the area we are looking for

So P (b) = 0.007  or 7 %

Finally between the two points above mentioned ( 10 ≤  Z  ≤ 12 ) we use the previous values (taking in consideration the limits, according to the problem statement )

Z ≤ 10     Z ⇒( 10-9  ) ÷ 1.2   Z = 0.7967

Z ≥ 12     Z ⇒ ( 12 - 9 ) ÷ 1.2  Z = 0.9952

The interval is between these two points

0.9952 - 0.7967 = 0.1985   ⇒ or 19.85 %

The attached help in the understanding of the solution

                               

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Zanzabum

Answer + step-by-step explanation:

7/10 as a decimal would be 0.7.

7/10 or 0.7 in a word form would be seven tenths.

<em><u>Hope this helps, if my answer or explanation is wrong  </u></em>

<em><u>feel free to message me in the comments :).</u></em>  

- genius423

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3 years ago
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Akimi4 [234]

Answer:

5+23

Step-by-step explanation:

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7 0
3 years ago
Read 2 more answers
Could someone help with the questions in the images below? I found them difficult
zubka84 [21]
Question 1a - You have got this correct, the median mark is 35
Question 1b- To work out the range, you must do the largest value subtract the smallest value. For your data, this would be 57 - 13 = 44

Question 2 - You have drawn the lines in the correct places, but you have not used a ruler. In order to get full marks on a question like this you must use a ruler.

Question 3 - Your lower quartile and median is correct, to find the upper quartile, we do 3(n/4). 'n' is the number of data points - in this case 120.
120/4 = 30
30 x 3 = 90
Go across the graph at 90 to see when the line is hit.
From what I can tell, the Upper Quartile is 3.

Next just find the maximum value.
Again, by the looks of it, the Maximum value is 8.5.

Now you have all the data needed for the box plot.

Min = 0
LQ = 0.8
Med = 2.1
UQ =  3
Max = 8.5

Using this information, draw a box plot like you did on the previous question.
<em>Again, I must stress that any line that you draw (unless purposely curved) must be drawn with a ruler; even on the graph at the top, during you working out. If you do not use a ruler, marks can be lost/taken away.</em>

Question 4a - For the median on a cumulative frequency chart, you must find the halfway point in the data. For this, we do n/2. In this case, n = 40
40/2 = 20
Draw a line (using  ruler) across at 20 until you reach the line.
Draw another one down to help you read the number.
The median looks like 34 seconds.
Question 4b - For this question, we already know the Min, Med and Max. Now we must work out the LQ and UQ.

Remember, LQ = n/4
In this case, n = 40
40/4 = 10
Look across at ten and you will find that the LQ = 16 seconds

To work out the UQ, we multiply the LQ place by 3 3(n/4).
3 x 10 = 30
30 on the graph takes us up to 45 seconds.

We now know that the:
Min = 9
LQ = 16
Med = 34
UQ = 45
Max = 57

With this information, draw a box plot like you have in the previous questions.

Question 5 - Good things to always compare on box plots are the Min/Max/Range and the Inter Quartile Range.
The boys had a lowest minimum and a higher maximum, this means that their range is larger, resulting in a large spread of data in comparison to the girls.
The Inter Quartile Range is the difference between the Upper Quartile and the Lower Quartile (how wide the box is).
The boy IQR = 45 - 16 = 29
The girls IQR = 34 - 23 = 11
Again, the girls have much more concise results; even though they did not get the quickest result, they were more like one another.

Question 5a - The median in a box plot is always the line down the centre of the box. In this case:
Median = 24 marks
Question 5b - The IQR is always the UQ - LQ
With this data, the IQR is the following:
36 - 17 = 19 marks

Hope this helps
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Reika [66]

Answer:

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Step-by-step explanation:

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2 years ago
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Vera_Pavlovna [14]

Answer:

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Step-by-step explanation:

Given : f(x) = x² + 5x + 6.

To find : What are the zeroes of f(x).

Solution : We have given that

f(x) = x² + 5x + 6.

To find the zeros of the function we need to set f(x) = 0.

x² + 5x + 6 = 0.

On factoring

x² + 3x +2x + 6 = 0.

Taking common x from first two term and 2 from last two terms

x ( x + 3) +2 ( x +3) = 0.

On grouping

(x + 3) (x + 2)  = 0

Now, x + 3 = 0   and x + 2 = 0

x = -3 and x = -2

Therefore, Zeros are x = −2, −3 .

3 0
3 years ago
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