HELPPPPPP PLEASEEEEE

A company rounds its losses to the nearest dollar. The error on each loss is independently and uniformly distributed on [–0.5, 0

lesya [120]

Answer:

the 95th percentile for the sum of the rounding errors is 21.236

Step-by-step explanation:

Let consider X to be the rounding errors

Then; $X \sim U (a,b)$

where;

a = -0.5 and b = 0.5

Also;

Since The error on each loss is independently and uniformly distributed

Then;

$\sum X _1 \sim N ( n \mu , n \sigma^2)$

where;

n = 2000

Mean $\mu = \dfrac{a+b}{2}$

$\mu = \dfrac{-0.5+0.5}{2}$

$\mu =0$

$\sigma^2 = \dfrac{(b-a)^2}{12}$

$\sigma^2 = \dfrac{(0.5-(-0.5))^2}{12}$

$\sigma^2 = \dfrac{(0.5+0.5)^2}{12}$

$\sigma^2 = \dfrac{(1.0)^2}{12}$

$\sigma^2 = \dfrac{1}{12}$

Recall:

$\sum X _1 \sim N ( n \mu , n \sigma^2)$

$n\mu = 2000 \times 0 = 0$

$n \sigma^2 = 2000 \times \dfrac{1}{12} = \dfrac{2000}{12}$

For 95th percentile or below

$P(\overline X < 95}) = P(\dfrac{\overline X - \mu }{\sqrt{{n \sigma^2}}}< \dfrac{P_{95}- 0 } {\sqrt{\dfrac{2000}{12}}}) =0.95$

$P(Z< \dfrac{P_{95} } {\sqrt{\dfrac{2000}{12}}}) = 0.95$

$P(Z< \dfrac{P_{95}\sqrt{12} } {\sqrt{{2000}}}) = 0.95$

$\dfrac{P_{95}\sqrt{12} } {\sqrt{{2000}}} =1- 0.95$

$\dfrac{P_{95}\sqrt{12} } {\sqrt{{2000}}} = 0.05$

From Normal table; Z > 1.645 = 0.05

$\dfrac{P_{95}\sqrt{12} } {\sqrt{{2000}}} =1.645$

${P_{95}\sqrt{12} } = 1.645 \times {\sqrt{{2000}}}$

${P_{95} = \dfrac{1.645 \times {\sqrt{{2000}}} }{\sqrt{12} } }$

$\mathbf{P_{95} = 21.236}$

the 95th percentile for the sum of the rounding errors is 21.236

8 0

3 years ago

Please solve the first three question with explanation

Lilit [14]

Answer:

first one is b, second one is c

Step-by-step explanation:

3-4x<=11

-4x<=11-3

-4x<=8/:(-4)

x>=-2

6 0

3 years ago

What is the completely factored form of d4 - 8d2 + 16?. A) (d2 + 4)(d2 - 4). B) (d2 - 4)(d2 - 4). C) (d2 + 4)(d + 2)(d - 2). D)

Wittaler [7]

For a quadratic equation, the negative of the numerical coefficient of the second term is the sum of the roots and the constant is their product. Let a and b be the roots such that,
a + b = 8
ab = 16
The values of a and b are 4 and 4. Such that the factors are,
(d² - 4)(d² -4)
Both the factors are difference of two squares. The final answer for this item is,
(d + 2)(d - 2)(d + 2)(d - 2)

This is letter D.

8 0

4 years ago

Over which interval does f have an average rate of change zero?

kvv77 [185]

Answer:

Step-by-step explanation:

The average rate of change is the slope. Slope has a formula that is the change in y over the change in x, which is a fraction. The only time a fraction can have a vlue of 0 is where the numerator of the fraction is equal to 0 (since we are not allowed to have a denominator of 0). If the change in y is in the top of the slope fraction, then we have to find the interval where the y values are the same. I'll show you one where the y values are not the same so you can compare it to the slope where the y values are the same. We will find the slope of choice A.

When x = -3, y = 0 so the coordinate is (-3, 0).

When x = 5, y = 5 so the coordinate is (5,4). Now let's find the slope (aka average rate of change) between those 2 coordinates:

$m=\frac{4-0}{5-(-3)}=\frac{4}{8}=\frac{1}{2}$ and the top of the fraction is a 1, not a 0, so the average rate of change between these 2 points is 1/2, not 0. Now let's do D.

When x = -3, y = 0 so the coordinate is (-3, 0).

When x = -1, y = 0 so the coordinate is (-1, 0). The slope between these 2 points is

$m=\frac{0-0}{-1-(-3)}=\frac{0}{2}=0$ This fraction is equal to 0 because the numerator is 0. Choice D is the one you want.

5 0

3 years ago

"a dealer in a casino has rolled a 6 on a single die four times in a row. what is the probability of his rolling another 6 on th

jeka94

One out of six or 1/6

4 0

3 years ago