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3 years ago

A random variable r is normally distributed with a mean of 7 and a standard deviation of 1.5. Find the value of w so that P (8.8

Mathematics

1 answer:

a_sh-v [17]3 years ago

4 0

Answer:

The answer is explained below

Step-by-step explanation:

Given that:

mean (μ) = 7 and standard deviation (σ) = 1.5.

The z score is used in statistic used to measure the number of standard deviation by which the raw score is above or below the mean value. It is given by:

$z=\frac{x-\mu}{\sigma}, where\ x\ is\ the \ raw\ score$

To find the Probability that x < 8.8, we first find the z score using:

$z = \frac{x-\mu}{\sigma}=\frac{8.8-7}{1.5}=1.2$

From the z tables, P(x < 8.8) = P(z < 1.2) = 0.8849 = 88.49%

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A highway had a landslide, where 3,000 cubic yards of material fell on the road, requiring 200 dump truck loads to clear. On ano

Orlov [11]

3,000 yd3 = 200 dtl
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12 A highway had a landslide, where 3,000 cubic yards of material fell on the road, requiring 200 dump truck loads to clear. On another highway, a slide left 40,000 cubic yards on the road. How many dump truck loads would be needed to clear this slide?
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3 years ago

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Hi Mrsdiaz1jd1977owwfvt!
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Suppose that, after measuring the duration of many telephone calls, a telephone company found their data was well-approximated b

Musya8 [376]

Answer:

a) 7.79%

b) 67.03%

c) Cumulative Distribution Function

$P(t) = \displaystyle\int^{\infty}_{-\infty} 0.1e^{-0.1t}~dt\\\\= \displaystyle\int^{b}_{a} 0.1e^{-0.1t}~dt, ~~a\leq t \leq b$

Step-by-step explanation:

We are given the following in the question:

$p(x) = 0.1 e^{-0.1x}$

where x is the duration of a call, in minutes.

a) P( calls last between 2 and 3 minutes)

$=\displaystyle\int^3_2 p(x)~ dx\\\\= \displaystyle\int^3_20.1e^{-0.1x}~dx\\\\=\Big[-e^{-0.1x}\Big]^3_2\\\\=-\Big[e^{-0.3}-e^{-0.2}\Big]\\\\= 0.0779\\=7.79\%$

b) P(calls last 4 minutes or more)

$=\displaystyle\int^{\infty}_4 p(x)~ dx\\\\= \displaystyle\int^{\infty}_40.1e^{-0.1x}~dx\\\\=\Big[-e^{-0.1x}\Big]^{\infty}_4\\\\=-\Big[e^{\infty}-e^{-0.4}\Big]\\\\=-(0- 0.6703)\\= 0.6703\\=67.03\%$

c) cumulative distribution function

$P(t) = \displaystyle\int^{\infty}_{-\infty} 0.1e^{-0.1t}~dt\\\\= \displaystyle\int^{b}_{a} 0.1e^{-0.1t}~dt, ~~a\leq t \leq b$

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3 years ago