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azamat
4 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]4 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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<span>y=x(x+5)(x-8)

The zeros of  </span><span>y=x(x+5)(x-8), means to solve the value of x, when y =0

</span><span>y=x(x+5)(x-8) = 0
</span><span>
x(x+5)(x-8) = 0

x = 0  or  (x+5) = 0   or (x - 8) = 0


x = 0               x + 5 = 0             x - 8 = 0    
                       x  = 0 -5              x = 0 + 8
                       x = -5                  x = 8

Hence the zeros are  x = 0,  x = -5,  x = 8</span>
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3 years ago
A woman invested 3 times as much at 6% as she did at 3%. The total amount of interest she earns in 1 year from both accounts is
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882 in one
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3 years ago
About 5% of hourly paid workers in a regioner the prevailing minimum wage or less A grocery chain offers discount rates to compa
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Answer:

0

Step-by-step explanation:

Let X  to be a random variable that looks a binomial distribution which denoted the number of employees out of the 281 who earn the prevailing minimum wage or less

The sample size n = 281

The population parameter p = 5% = 0.05

Using normal approximation for the mean.

\mu = np

\mu = 281\times 0.05

\mu = 14.05

The standard deviation is:

\sigma = \sqrt{np(1-p)}

\sigma = \sqrt{281 \times 0.05(1-0.05)}

\sigma = \sqrt{281 \times 0.05(0.95)}

\sigma = \sqrt{13.3475}

\sigma =3.6534

By using continuity correction; the sample mean x is:

x = 30 - 0.5

x = 29.5

The z statistic test can now be as follows:

Z = \dfrac{x-\mu}{\sigma}

Z = \dfrac{29.5-14.05}{3.6534}

Z = \dfrac{15.45}{3.6534}

Z = 4.23

Thus, the probability that company A will get a discount is

P(X ≥ 30) = P(Z >4.23)

= 1 - P(Z < 4.23)

By using the Excel function for the z score 4.23 i.e. "=1 - NORMSDIST(4.23)" we get;

= 0.0000

7 0
3 years ago
Math question down below
amm1812

Answer:

D. 604

Step-by-step explanation:

411/3 = 137

137 + 388 = 525

----------------------------------

(n) 604 - 79 = 525

8 0
3 years ago
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