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andreev551 [17]
3 years ago
7

At a large business, employees must report to work at 7:30 A.M. The arrival times of employees is approximately symmetric and mo

und-shaped with mean 7:22 A.M. and standard deviation 4 minutes.
Question 1. Use the 68-95-99.7 rule (also known as the Empirical rule) to determine what percent of employees are late on a typical day.

Question 2. A psychological study determined that the typical worker needs four minutes to adjust to their surroundings before beginning their duties. Use the 68-95-99.7 rule (also known as the Empirical rule) to determine what percent of this business? employees arrive early enough to make this adjustment before the 7:30 A.M. start time.
Mathematics
1 answer:
Hatshy [7]3 years ago
8 0

Answer:

(a) 2.5%

(b) 84%

Step-by-step explanation:

Applying the empirical rule,

68% arrive within 7:22 am +/- 4 minutes (7:18 am to 7:26 am)

95% arrive within 7:22 am +/- 8 minutes (7:14 am to 7:30 am)

99.7% arrive within 7:22 am +/- 12 minutes (7:10 am to 7:34 am)

(a) Those that arrive late arrive after 7:30 am. Half of all employees (50%) arrive after 7:22 am.

Since the arrival time is symmetric, half of those that arrive between 7:14 am and 7:30 am (95%) arrive between 7:22 am and 7:30 am.

This means 95% ÷ 2 = 47.5% arrive before 7:30 am.

The late arrivals = 50% - 47.5% = 2.5%

(b) Those who come early to make the adjustment come 4 minutes before 7:30 am = 7:26 am.

50% come before 7:22 am.

Since the arrival time is symmetric, half of those that arrive between 7:18 am and 7:26 am (68%) arrive between 7:22 am and 7:26 am. This means 68% ÷ 2 = 34% arrive between 7:22 am and 7:26 am.

The percentage that arrive before 7:26 am = 50% + 34% = 84%.

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The sum of the given series can be found by simplification of the number

of terms in the series.

  • 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

Which gives, the number of terms, n = 2020

\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}

  • A ≈ <u>2020.022</u>

Learn more about the sum of a series here:

brainly.com/question/190295

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