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2 years ago

A normal population has a mean of 20.0 and a standard deviation of 4.0.

1 answer:

3 0

<h2><em>a) </em><u>The z-value associated with 25.0</u></h2><h2>Z = 1.25</h2><h2 /><h2><em>b)</em> <u>proportion of the population is between 20.0 and 25.0 is</u></h2><h2>Proportion = 0.3944</h2><h2 /><h2><em>c)</em> <u>proportion of the population is less than 18.0 is</u></h2><h2>Proportion = 0.3085</h2>

Calculations:

Normal pop has mean of 20.0

standard deviation = 4.0

XNN(20.0, 4.0)

a).

$z=\frac{x-h}{z}$

$=\frac{25-20}{4.0} =\frac{5}{4} =1.25$

<u></u>

b).

The proportion between 20 and 25 is P(20 <x<25.0)

$=p(\frac{20-20}{4} < z < \frac{25-0}{4} )$

$=P(0 < z < 1.25)$

$=P(Z < 1.25)-P(z < 0)$

$=0.8944-0.5000$

P(20 < x < 25)=0.3944

c).

The proportion value is less than 18 when

$P(x < 18)=p(\frac{x--4}{6} < \frac{18-20}{4}$

$=P(z < \frac{-2}{4})$

$=P(z < -0.5)$

P(x<18) = 0.3085

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A certain chain of ice cream stores sells 28 different flavors, and a customer can order a single-, double- , or triple-scoop co

astra-53 [7]

<span>Think about the cones with only 1 scoop of ice cream. Isn't it clear that there are 28 of those?
Now think about the two scoop cones. You have 28 choices for the first scoop and 28 choices for the second scoop.
So the number of possibilities is 28(28).
Now think of the 3 scoop cones.
You have 28 choices for the first scoop, 28 choices for the second scoop, and 28 choices for the third scoop or 28(28)(28) possibilities.
Add them all together and you have the total.

So it will be like this:
</span><span>28^3+28^2+28
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I hope my answer helped you.

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3 years ago

I need hlep its do to day

Taya2010 [7]

Answer:

x = -6/5

Step-by-step explanation:

distribute -4 on the left side of the equation first, then distribute 2 on the right side of the equation to get:

16x - 4 - 4x = 2x - 14 - 2

combine 'like terms':

12x - 4 = 2x - 16

subtract '2x' from each side to get:

10x - 4 = -16

add 4 to each side to get:

10x = -12

x = -12/10 or x = -6/5

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3 years ago

Which of the following fall(s) at a z-score of 0.0 in a normal distribution? Mean Median Mode Mean, media and mode

Artist 52 [7]

Answer: Median Mode Mean

Step-by-step explanation:

A normal distribution a symmetric distribution where most of the values lies around the central peak where Mean, median mode all lies together and the value of z=0 (as z-value given the distance of data values from mean with respect to the standard deviation).

i.e. Z-score of 0 has 50% of the area to the left and 50% area to the right.

Hence, at a z-score of 0.0 in a normal distribution Mean, Median and Mode all fall together.

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3 years ago

The number of people arriving at a ballpark is random, with a Poisson distributed arrival. If the mean number of arrivals is 10,

Stella [2.4K]

Answer:

a) 3.47% probability that there will be exactly 15 arrivals.

b) 58.31% probability that there are no more than 10 arrivals.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given time interval.

If the mean number of arrivals is 10

This means that $\mu = 10$

(a) that there will be exactly 15 arrivals?

This is P(X = 15). So

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 15) = \frac{e^{-10}*(10)^{15}}{(15)!} = 0.0347$

3.47% probability that there will be exactly 15 arrivals.

(b) no more than 10 arrivals?

This is $P(X \leq 10)$

$P(X \leq 10) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)$

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-10}*(10)^{0}}{(0)!} = 0.000045$

$P(X = 1) = \frac{e^{-10}*(10)^{1}}{(1)!} = 0.00045$

$P(X = 2) = \frac{e^{-10}*(10)^{2}}{(2)!} = 0.0023$

$P(X = 3) = \frac{e^{-10}*(10)^{3}}{(3)!} = 0.0076$

$P(X = 4) = \frac{e^{-10}*(10)^{4}}{(4)!} = 0.0189$

$P(X = 5) = \frac{e^{-10}*(10)^{5}}{(5)!} = 0.0378$

$P(X = 6) = \frac{e^{-10}*(10)^{6}}{(6)!} = 0.0631$

$P(X = 7) = \frac{e^{-10}*(10)^{7}}{(7)!} = 0.0901$

$P(X = 8) = \frac{e^{-10}*(10)^{8}}{(8)!} = 0.1126$

$P(X = 9) = \frac{e^{-10}*(10)^{9}}{(9)!} = 0.1251$

$P(X = 10) = \frac{e^{-10}*(10)^{10}}{(10)!} = 0.1251$

$P(X \leq 10) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 0.000045 + 0.00045 + 0.0023 + 0.0076 + 0.0189 + 0.0378 + 0.0631 + 0.0901 + 0.1126 + 0.1251 + 0.1251 = 0.5831$

58.31% probability that there are no more than 10 arrivals.

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3 years ago

A police officer investigating an accident finds a skid mark of 115 ft in length. How fast was the car going when the driver app

Dominik [7]

The question doesn't provide enough information to get an accurate answer. Please provide more information or provide acceptable assumptions.

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3 years ago