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3 years ago

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An advisor to the mayor of a large city wants to estimate, within 3 minutes, the mean travel time to work for all employees who

work within the city limits. He knows that the standard deviation for all travel times is 12.25 minutes. He also wants to achieve a 95% confidence interval. He will poll a random sample of city employees. At least how many employees should he poll?

1 answer:

Shkiper50 [21]3 years ago

8 0

Answer:

64

Step-by-step explanation:

Given that an advisor to the mayor of a large city wants to estimate, within 3 minutes, the mean travel time to work for all employees who work within the city limits.

Let X be the random variable denoting the time for employees to travel within city limit

Std dev of X $= 12.25 minutes$

For 95% confidence interval Z critical 1.96 is used because population std dev is known

Margin of error $= 1.96*std error = 3 (given)$

i.e. $1.96(\frac{12.25}{\sqrt{n} } )=3$

where n is the desired sample size.

Simplify to get

$\frac{24.01}{\sqrt{n} } =3\\n=64.053$

Hence atleast 64 employees should he poll

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<h3>What is scientific notation?</h3>

Extremely big or small numbers are presented in a more straightforward manner using scientific notation. When writing numbers that are frequently too long to type or comprehend, many scientists and engineers utilize scientific notation. When a number is written in scientific notation, it becomes a decimal rather than a long string of integers that might not be necessary for the equation as a whole. The number is now easier to comprehend and enter into an equation thanks to the new decimal value. The following is an example of scientific notation.

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Scientific notation is a special case of “exponential notation.” Exponential notation has three parts - a base, an exponent, and a “significand.” If you let the base be “d”, the exponent be n, and the significand be c, you’d write a number like this:

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We have here a <em>normally distributed random variable</em>. The parameters that characterize this distribution is the <em>mean</em>, $\\ \mu$ , and the <em>standard deviation</em>, $\\ \sigma$ .

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And we want to know the probability that a randomly selected single egg "will be with cholesterol content greater than 220 milligrams."

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Z-scores.
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The <em>z-scores</em> are standardized values and represent the distance (for the raw score) from the mean in standard deviations units. A <em>positive</em> z-score indicates that it is <em>above</em> $\\ \mu$ and, conversely, a negative result that the value is <em>below</em> it.

The <em>cumulative distribution function</em> generates the values for the <em>cumulative standard normal distribution</em> displayed in the <em>standard normal table</em>.

The <em>standard normal distribution</em> is employed to find probabilities for any normally distributed data and we only need to calculate the corresponding z-score (or standardized value). This distribution has a $\\ \mu = 0$ and $\\ \sigma = 1$ .

As we can see, all of these concepts are intertwined, and each of them is important because:

To find the corresponding probability, we first need to obtain the <em>z-score</em>.
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Finding the probability

We can get the<em> z-score</em> using the formula:

$\\ z = \frac{x - \mu}{\sigma}$ [1]

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Thus (without using units):

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$\\ P(x$

However, the question is asking for "cholesterol content greater than 220 milligrams" or

$\\ P(x>220) = P(z>0.33)$

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$\\ P(x220) = 1$

Which is the same for a standardized value:

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Therefore

$\\ P(x>220) = P(z>0.33) = 1 - P(z$

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The graph below shows a shaded area that corresponds to the found probability.

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