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xenn [34]
3 years ago
11

The record high January temperature in Austin, Texas is 90 degres F. The record low January temperature is -2 degres F. Find the

difference between the high and the low temperatures
Mathematics
2 answers:
Elan Coil [88]3 years ago
7 0
For this one it's really simple. Just do 90-(-2). It may look confusing right now, but it's really easy. All you need to know is that a negative and a negative cancel out to form a positive. So it's actually 90+2. 92 is the difference.
Nostrana [21]3 years ago
5 0
Its really simple when two negative operation are next to each other that turn into a plus so I think 94 is the answer
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Answer:

a) P(X \leq 4) = 0.6289

P(X < 4) = 0.4335

b) P(4 \leq X \leq 8) = 0.5452

c) P(X \geq 8) = 0.0511

d) 0.2605

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.

It is important to know that the variance has the same value as the mean in a Poisson distribution. The standard deviation is the square root of the variance.

In this problem, we have that:

\mu = 4, \sigma = \sqrt{4} = 2.

To help our solution, i am going to find each of P(X = x) from x = 0 to 8[/tex]

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

P(X = 0) = \frac{e^{-4}*4^{0}}{(0)!} = 0.0183

P(X = 1) = \frac{e^{-4}*4^{1}}{(1)!} = 0.0733

P(X = 2) = \frac{e^{-4}*4^{2}}{(2)!} = 0.1465

P(X = 3) = \frac{e^{-4}*4^{3}}{(3)!} = 0.1954

P(X = 4) = \frac{e^{-4}*4^{4}}{(4)!} = 0.1954

P(X = 5) = \frac{e^{-4}*4^{5}}{(5)!} = 0.1563

P(X = 6) = \frac{e^{-4}*4^{6}}{(6)!} = 0.1042

P(X = 7) = \frac{e^{-4}*4^{7}}{(7)!} = 0.0595

P(X = 8) = \frac{e^{-4}*4^{8}}{(8)!} = 0.0298

(a) Compute both P(X ≤ 4) and P(X < 4)

P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0183 + 0.0733 + 0.1465 + 0.1954 + 0.1954 = 0.6289

---------

P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0183 + 0.0733 + 0.1465 + 0.1954 = 0.4335

(b) Compute P(4 ≤ X ≤ 8).

P(4 \leq X \leq 8) = P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) = 0.1954 + 0.1563 + 0.1042 + 0.0595 + 0.0298 = 0.5452

(c) Compute P(8 ≤ X)

That is P(X \geq 8).

The sum of the probabilities must be 1 in decimal. Either X is greater or equal to 8, or X is lesser than 8.

So

P(X < 8) + P(X \geq 8) = 1

P(X \geq 8) = 1 - P(X < 8)

From what we have in a) and b)

P(X < 8) = P(X < 4) + P(4 \leq X < 8) = 0.4335 + 0.1954 + 0.1563 + 0.1042 + 0.0595 = 0.9489

So

P(X \geq 8) = 1 - P(X < 8) = 1 - 0.9489 = 0.0511

(d) What is the probability that the number of anomalies exceeds its mean value by no more than one standard deviation

One standard deviation is 2, and the mean is 4. So, this is:

P(4 < X \leq 6) = P(X = 5) + P(X = 6) =  0.1563 + 0.1042 = 0.2605

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3 years ago
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