Juanita is converting 7 meters to inches. Will the number of inches be less than or greater than 7?

High school students across the nation compete in a financial capability challenge each year by taking a National Financial Capa

Nikolay [14]

Using the normal distribution, it is found that a student has to score 1.08 standard deviations above the mean to be publicly recognized.

<h3>Normal Probability Distribution</h3>

In a normal distribution with mean $\mu$ standard deviation $\sigma$ z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

It measures how many standard deviations the measure is from the mean.

After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

The top 14% of the scores are given by scores above the 86th percentile, which corresponds to Z = 1.08, hence, a student has to score 1.08 standard deviations above the mean to be publicly recognized.

More can be learned about the normal distribution at brainly.com/question/24663213

4 0

3 years ago

Speedy Swift is a package delivery service that serves the greater Atlanta, Georgia, metropolitan area. To maintain customer loy

Paladinen [302]

Answer:

See Explanation and attachments

Step-by-step explanation:

Given

See question for data

Solving (a):

The scale used is an ordinal scale.

Ordinals scale uses hierarchy arrangement and the data can be arranged as:

$EARLY ->> ON-TIME ->> LATE ->> LOST$

The variable of the delivery performance is qualitative because it is non-numerical

Solving (b): Frequency table

From the table, we have:

$On - Time = 57$ $Early = 19$ $Late = 9$ $Lost = 2$

So, the frequency distribution table is:

$\begin{array}{cc}{Performance}&{Frequency}&{On-Time}&{57}&{Early}&{19}&{Late}&{9} & {Loss} & {2} & {Total} & {87}\end{array}$

Solving (c): Frequency table.

To do this, we add another column (Relative Frequency) to the above table.

The relative frequency is calculated as:

$Relative\ Frequency = \frac{Frequency}{Total}$

So, we have:

$On - Time = \frac{57}{87} = 0.656$ $Early = \frac{19}{87} = 0.218$

$Late = \frac{9}{87} = 0.103$ $Lost = \frac{2}{87} = 0.023$

So, the frequency distribution table is:

$\begin{array}{ccc}{Performance}&{Frequency}&{Relative\ Frequency} & {On-Time}&{57}&{0.656}&{Early}&{19}&{0.218} & {Late}&{9} & {0.103} & {Loss} & {2} & {0.023} & {Total} & {87}&{1}\end{array}$

Solving (d & e): See attachment 1 for bar chart & attachment 2 for pie chart

Solving (f):

From the question, we understand that the object is to return 99% early or on time and never to lose a package.

The analysis is as follows:

Early and On-Time (ET) packages

$ET = \frac{Early + On-Time}{Total} * 100\%$

$ET = \frac{57 + 19}{87} * 100\%$

$ET = \frac{76}{87} * 100\%$

$ET = \frac{7600}{87} \%$

$ET = 87.4 \%$

Lost packages

$Lost = \frac{Lost}{Total} * 100\%$

$Lost = \frac{2}{87} * 100\%$

$Lost = \frac{200}{87} \%$

$Lost = 2.30\%$

<em>From the above analysis, we can see that 87.4% of the packages were delivered early enough and 2.30% were lost. </em>

<em>The fraction of packages delivered early can be improved and the fraction of lost packages can be reduced by exploring the chances of taking alternative routes when possible. </em>