Jamila has to purchase a new car and is putting

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

Calculate the area of the shapes below. 2. 8 ft 4 ft 12 ft Please LOL

Kay [80]

Answer:

good luck!

Step-by-step explanation:

3 0

3 years ago

How do you simplify 55/20

nasty-shy [4]

Find the greatest common factor. For this it would be 11/4

3 0

3 years ago

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Players on the school soccer team are selling candles to raise money for an upcoming trip. Each player has 24 candies to sell. I

liubo4ka [24]

Answer:

Slope, M = 5 and this means that profit is five (5) times every unit change in the quantity of candies.

Step-by-step explanation:

Let the equation below represent the Profit and number of candies relationship:

π (x) = Mx +c--------------------------------------------------------------- (1)

Where π = Profit

x = Number of candies sold

M = Slope of the linear relationship/ equation

C = The profit intercept of the linear equation or relationship

From the question two different coordinates of profit and the number of candies where given: (π₁ ,x₁) and (π₂ ,x₂)

First coordinate (π₁ ,x₁) = ($30 , 4)

Second coordinate (π₂ ,x₂) = ($70, 12)

These can be substituted into equation (1) and (2) to calculate the for M & C

Substituting the first coordinate into (1) we have :

30 = 4M +C-----------------------------------------------------------------(2)

Substituting the second coordinate into (1) we have :

70 = 12M +C------------------------------------------------------------------(3)

Solving equation (2) and (3) simultaneously using elimination method, we have:

30 = 4M +C

70 = 12M +C

-40 = -8M

M = 5

Substituting the value of M into equation (3) we have:

70 = 12(5) +C

70 = 60 +C

C =10

Substituting the value of M and C into equation (1), we have the Linear relationship for profit and the number of candies sold as

π (x) = 5x + 10---------------------------------------------------------------- (4)

Slope = 5 and this means that profit changes five (5) times for every unit change in the quantity of candies.

5 0

3 years ago

Anna drew a scale drawing of the elementary school. The school yard, which is 142 meters long in real life, is 213 millimeters l

Paraphin [41]

Answer:

30,246

Step-by-step explanation:

multiply the two numbers

6 0

3 years ago

Read 2 more answers