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azamat
3 years ago
13

In a normally distributed data set a mean of 55 where 99.7% of the data fall between 47.5 and 62.5, what would be the standard d

eviation of that data set? Devry course
Mathematics
1 answer:
NeX [460]3 years ago
3 0

Given that normally distributed data set has a mean of 55 and 99.7% of data fall between 47.5 and 62.5.

Let s be the standard deviation of data set.

Since 99.7% data fall within 3 standard deviations of mean, z-value for 47.5 and 62.5 has an absolute value of 3.

That is |z|=3

But z= \frac{x-mean}{standard deviation}

Let us plugin x=47.5 and mean =55 and equate it to 3.

That is |\frac{47.5-55}{s}|  = 3

             |\frac{-7.5}{s} | =3

Since x is always positive ( being standard deviation), |\frac{-7.5}{s} | = \frac{|-7.5|}{s} = \frac{7.5}{s}

Hence  \frac{7.5}{s}= 3

             s=\frac{7.5}{3}  = 2.5

We will get same value with 62.5 as well.

Hence standard deviation of data set is 2.5.

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Answer:

9 sides.

Step-by-step explanation:

N is the # of sides.

140 = ((n-2) x 180) / n

Multiply both sides by n

140 n = (n-2) x 180

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140 n = 180 n - 360

Subtract 140 n and add 360 to both sides.

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The value of the expression 10 - 1/2^4 x 48<br> A = 2<br> B = 4<br> C = 5<br> D = 7
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Answer:

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Polly surveyed 850 teenagers to find out their favorite type of music. She found that 32% of the teenagers surveyed like hard ro
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Answer:

In explanation

Step-by-step explanation:

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2 years ago
G(-4)=3(-4)+4<br><br> Find g(-4)
soldier1979 [14.2K]
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3 years ago
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

6 0
3 years ago
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