A normal probability/quantile plot is used to see if the distribution of a quantitative variable follows a __________ distributi
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The sparkling-water distributor should use 160 gallons of $ 14 and 80 gallons of $ 3 and 160 gallons of $ 4.50
<em><u>Solution:</u></em>
Let "x" be the gallons of $ 14
Let "y" be the gallons of $ 3
Let "z" be the gallons of $ 4.5
<em><u>A sparkling-water distributor wants to make up 400 gal of sparkling water</u></em>
Therefore,
x + y + z = 400 --------- eqn 1
<em><u>She must use twice as much of the $4.50 water as the $3.00 water</u></em>
gallons of $ 4.5 = twice of gallons of $ 3
z = 2y --------- eqn 2
<em><u>A sparkling-water distributor wants to make up 400 gal of sparkling water to sell for $8.00 per gallon</u></em>
Therefore, we frame a equation as:
14x + 3y + 4.5z = 3200 ----------- eqn 3
<em><u>Substitute eqn 2 in eqn 3</u></em>
14x + 3y + 4.5(2y) = 3200
14x + 3y + 9y = 3200
14x + 12y = 3200
Divide both sides by 2
7x + 6y = 1600 -------- eqn 4
<em><u>Substitute eqn 2 in eqn 1</u></em>
x + y + 2y = 400
x + 3y = 400
x = 400 - 3y ------- eqn 5
<em><u>Substitute eqn 5 in eqn 4</u></em>
7(400 - 3y) + 6y = 1600
2800 - 21y + 6y = 1600
15y = 1200
Divide both sides by 15
y = 80
<em><u>Substitute y = 80 in eqn 5</u></em>
x = 400 - 3(80)
x = 400 - 240
x = 160
<em><u>Substitute y = 80 in eqn 2</u></em>
z = 2(80)
z = 160
Thus she should use 160 gallons of $ 14 and 80 gallons of $ 3 and 160 gallons of $ 4.50
Answer:
0.71%
Step-by-step explanation:
Given that a normally distributed set of data has a mean of 102 and a standard deviation of 20.
Let X be the random variable
Then X is N(102, 20)
We can convert this into standard Z score by
We are to find the probability and after wards percentage of scores in the data set expected to be below a score of 151.
First let us find out probability using std normal table
P(X<151) =
We can convert this into percent as muliplying by 100
percent of scores in the data set expected to be below a score of 151.
=0.71%