1-A) Refer to Section 4.1 Exercise 8 in the text. Does the curve represent a con

Question

1-A)

Refer to Section 4.1 Exercise 8 in the text. Does the curve represent a continuous probability distribution? If so, identify the type of distribution.

B)  

Fill in the blank to complete the uniform distribution that is described.

A uniform distribution's density curve is defined by the horizontal line y = 0.25 starting at x = -2 and ending at x = ___.

C)

For the following normal distribution, give the x-values of the inflection points of the curve (the points where the curve's concavity changes).

x ~ N(0, 52)

D)

For the following normal distribution, give the x-values of the inflection points of the curve (the points where the curve's concavity changes).

x ~ N(-5, 1.252)

-7.5 and -2.5

Please help

a. Not a continuous probability distribution b. Other continuous probability distribution c. Uniform probability distribution d. Normal or Student t probability distribution

Explanation / Answer

B)

The area under the curve f(x) is = 1

0.25 * (x + 2) = 1

=> x = 2

Hence the answer is = (d)

For C & D:

A random variable that is normally distributed with mean ? and standard deviation of ? has a probability density function of

f( x ) =1/ (? ?(2 ?) )exp[-(x - ?)2/(2?2)].

Here we use the notation exp[y] = ey

The first derivative of this probability density function is found by knowing the derivative for ex and applying the chain rule.

f’ (x ) = -(x - ?)/ (?3 ?(2 ?) )exp[-(x -?) 2/(2?2)] = -(x - ?) f( x )/?2.

We now calculate the second derivative of this probability density function. We use the product rule to see that:

f’’( x ) = - f( x )/?2 - (x - ?) f’( x )/?2

Simplifying this expression we have

f’’( x ) = - f( x )/?2 + (x - ?)2 f( x )/(?4)

Now set this expression equal to zero and solve for x. Since f( x ) is a nonzero function we may divide both sides of the equation by this function.

0 = - 1/?2 + (x - ?)2 /?4

To eliminate the fractions we may multiply both sides by ?4

0 = - ?2 + (x - ?)2

We are now nearly at our goal. To solve for x we see that

?2 = (x - ?)2

By taking a square root of both sides (and remembering to take both the positive and negative values of the root

±? = x - ?

From this it is easy to see that the inflection points occur where x = ? ± ?. In other words the inflection points are located one standard deviation above the mean and one standard deviation below the mean.

C)

The inflection points are 0 - ?52 = - 7.21 and 0 + ?52 = 7.21

D)

The inflection points are - 5 - ?1.252 = - 6.119 and - 5 + ?1.252 = - 3.881

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