An accelerated life test on a large number of type-D alkaline batteries revealed that the mean life for a particular use before
2 answers:
Answer:
About 68.26% of the batteries failed between what two values?
About 95.44% of the batteries failed between what two values?
About 99.97% of the batteries failed between what two values?
14.7 and 23.3
Step-by-step explanation:
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
The empirical rule, also referred to as the three-sigma rule or 68-95-99.7 rule, is a statistical rule which states "that for a normal distribution, almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ). Broken down, the empirical rule shows that 68% falls within the first standard deviation (µ ± σ), 95% within the first two standard deviations (µ ± 2σ), and 99.7% within the first three standard deviations (µ ± 3σ)".
Let X the random variable who represent the mean life for a particular use before they failed.
From the problem we have the mean and the standard deviation for the random variable X.
Other way to interpret the empirical rule is like this:
• The probability of obtain values within one deviation from the mean is 0.68
• The probability of obtain values within two deviation's from the mean is 0.95
• The probability of obtain values within three deviation's from the mean is 0.997
About 68.26% of the batteries failed between what two values?
We can use the following codes in order to find the limits:
"=NORM.INV(0.1587,19,1.2)" and "=NORM.INV(1-0.1587,19,1.2)"
About 95.44% of the batteries failed between what two values?
We can use the following codes in order to find the limits:
"=NORM.INV(0.0228;19;1.2)" and "=NORM.INV(1-0.0228;19;1.2)"
About 99.97% of the batteries failed between what two values?
For this case the empirical rule is not the best approximation since within 3 deviations from the mean we have 99.7% of the data and not 99.97%. We can use excel to find the limits.
We can use the following codes in order to find the limits:
"=NORM.INV(0.00015;19;1.2)" and "=NORM.INV(1-0.00015;19;1.2)"
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case a)
the equation of the vertical parabola write in vertex form is
y=a(x-h)²+k,
where (h, k) is the vertex.
Using our vertex, we have:
y=a(x-2)²-1
We know that the parabola goes through (5, 0),
so
we can use these coordinates to find the value of a:
0=a(5-2)²-1
0=a(3)²-1
0=9a-1
Add 1 to both sides:
0+1=9a-1+1
1=9a
Divide both sides by 9:
1/9 = 9a/9
1/9 = a
y=(1/9)(x-2)²-1
the answer is
a=1/9
case b)
the equation of the horizontal parabola write in vertex form is
x=a(y-k)²+h,
where (h, k) is the vertex.
Using our vertex, we have:
x=a(y+1)²+2,
We know that the parabola goes through (5, 0),
so
we can use these coordinates to find the value of a:
5=a(0+1)²+2
5=a+2
a=5-2
a=3
x=3(y+1)²+2
the answer is
a=3
see the attached figure
case a)
the equation of the vertical parabola write in vertex form is
y=a(x-h)²+k,
where (h, k) is the vertex.
Using our vertex, we have:
y=a(x-2)²-1
We know that the parabola goes through (5, 0),
so
we can use these coordinates to find the value of a:
0=a(5-2)²-1
0=a(3)²-1
0=9a-1
Add 1 to both sides:
0+1=9a-1+1
1=9a
Divide both sides by 9:
1/9 = 9a/9
1/9 = a
y=(1/9)(x-2)²-1
the answer is
a=1/9
case b)
the equation of the horizontal parabola write in vertex form is
x=a(y-k)²+h,
where (h, k) is the vertex.
Using our vertex, we have:
x=a(y+1)²+2,
We know that the parabola goes through (5, 0),
so
we can use these coordinates to find the value of a:
5=a(0+1)²+2
5=a+2
a=5-2
a=3
x=3(y+1)²+2
the answer is
a=3
see the attached figure