Challenge Problem In statistics, the standard normal density function is given by f ( x ) = 1 2 π ⋅ exp [ − x 2 2 ] . This function can be transformed to describe any general normal distribution with mean μ , and standard deviation, σ . A general normal density function is given by f ( x ) = 1 2 π ⋅ σ ⋅ exp [ − ( x − μ ) 2 2 σ 2 ] . Describe the transformations needed to get from the graph of the standard normal function to the graph of a general normal function.
Challenge Problem In statistics, the standard normal density function is given by f ( x ) = 1 2 π ⋅ exp [ − x 2 2 ] . This function can be transformed to describe any general normal distribution with mean μ , and standard deviation, σ . A general normal density function is given by f ( x ) = 1 2 π ⋅ σ ⋅ exp [ − ( x − μ ) 2 2 σ 2 ] . Describe the transformations needed to get from the graph of the standard normal function to the graph of a general normal function.
Solution Summary: The author explains the transformations needed to graph the general normal function f(x)=1sqrt
Challenge Problem In statistics, the standard normal density function is given by
f
(
x
)
=
1
2
π
⋅
exp
[
−
x
2
2
]
. This function can be transformed to describe any general normal distribution with mean
μ
,
and standard deviation,
σ
. A general normal density function is given by
f
(
x
)
=
1
2
π
⋅
σ
⋅
exp
[
−
(
x
−
μ
)
2
2
σ
2
]
. Describe the transformations needed to get from the graph of the standard normal function to the graph of a general normal function.
Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.
Given lim x-4 f (x) = 1,limx-49 (x) = 10, and lim→-4 h (x) = -7 use the limit properties
to find lim→-4
1
[2h (x) — h(x) + 7 f(x)] :
-
h(x)+7f(x)
3
O DNE
17. Suppose we know that the graph below is the graph of a solution to dy/dt = f(t).
(a) How much of the slope field can
you sketch from this information?
[Hint: Note that the differential
equation depends only on t.]
(b) What can you say about the solu-
tion with y(0) = 2? (For example,
can you sketch the graph of this so-
lution?)
y(0) = 1
y
AN
(b) Find the (instantaneous) rate of change of y at x = 5.
In the previous part, we found the average rate of change for several intervals of decreasing size starting at x = 5. The instantaneous rate of
change of fat x = 5 is the limit of the average rate of change over the interval [x, x + h] as h approaches 0. This is given by the derivative in the
following limit.
lim
h→0
-
f(x + h) − f(x)
h
The first step to find this limit is to compute f(x + h). Recall that this means replacing the input variable x with the expression x + h in the rule
defining f.
f(x + h) = (x + h)² - 5(x+ h)
=
2xh+h2_
x² + 2xh + h² 5✔
-
5
)x - 5h
Step 4
-
The second step for finding the derivative of fat x is to find the difference f(x + h) − f(x).
-
f(x + h) f(x) =
= (x²
x² + 2xh + h² -
])-
=
2x
+ h² - 5h
])x-5h) - (x² - 5x)
=
]) (2x + h - 5)
Macbook Pro
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