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ZanzabumX [31]
2 years ago
7

The total snowfall per year in Laytonville is normally distributed with mean 99 inches and standard deviation 14 inches. Based o

n the Empirical Rule, what is the probability that in a randomly selected year, the snowfall was less than 127 inches
Mathematics
1 answer:
Sergio039 [100]2 years ago
3 0

Answer: Required probability is 0.9772.

Step-by-step explanation:

Since we have given that

Mean = 99 inches

Standard deviation = 14 inches

It is normally distributed.

We need to find the probability that in a randomly selected year, the snowfall was less than 127 inches.

As we know that

z=\dfrac{X-\mu}{\sigma}

So, it becomes,

P(X

Hence, required probability is 0.9772.

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A = 1011 + 337 + 337/2 +1011/10 + 337/5 + ... + 1/2021
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The sum of the given series can be found by simplification of the number

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  • A is approximately <u>2020.022</u>

Reasons:

The given sequence is presented as follows;

A = 1011 + 337 + 337/2 + 1011/10 + 337/5 + ... + 1/2021

Therefore;

  • \displaystyle A = \mathbf{1011 + \frac{1011}{3} + \frac{1011}{6} + \frac{1011}{10} + \frac{1011}{15} + ...+\frac{1}{2021}}

The n + 1 th term of the sequence, 1, 3, 6, 10, 15, ..., 2021 is given as follows;

  • \displaystyle a_{n+1} = \mathbf{\frac{n^2 + 3 \cdot n + 2}{2}}

Therefore, for the last term we have;

  • \displaystyle 2043231= \frac{n^2 + 3 \cdot n + 2}{2}

2 × 2043231 = n² + 3·n + 2

Which gives;

n² + 3·n + 2 - 2 × 2043231 = n² + 3·n - 4086460 = 0

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\displaystyle \frac{A}{2}  = \mathbf{ 1011 \cdot  \left(\frac{1}{2} +\frac{1}{6} + \frac{1}{12}+...+\frac{1}{4086460}  \right)}

\displaystyle \frac{A}{2}  = 1011 \cdot  \left(1 - \frac{1}{2} +\frac{1}{2} -  \frac{1}{3} + \frac{1}{3}- \frac{1}{4} +...+\frac{1}{2021}-\frac{1}{2022}  \right)

Which gives;

\displaystyle \frac{A}{2}  = 1011 \cdot  \left(1 - \frac{1}{2022}  \right)

\displaystyle  A = 2 \times 1011 \cdot  \left(1 - \frac{1}{2022}  \right) = \frac{1032231}{511} \approx \mathbf{2020.022}

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