Solve for x: 1.4x – 0.9 = 3.3

Distributions and Comparing Data Project

LenKa [72]

Answer:

1. The shape of the histogram created with MS Excel is approximately bell shaped and approximately evenly spread about the central (largest counts) values

The shape of the box plot created with MS Excel is approximately evenly distributed

2. The appropriate measure of central tendency for a bell shaped histogram is the mean and median, due to the approximately equal distribution about the highest frequency class

3. Part A

The 15 numbers between 1 and 20 generated by a random generator are;

1, 8, 8, 15, 4, 18, 11, 6, 17, 1, 18, 15, 10, 12, 11

The measure of center is the mean = (1+8+8+15+4+18+11+6+17+1+18+15+10+12+11)/15 = 155/15 = 31/3

The measure of spread used is the variance, s² = 32.38

$Where, \,s^2 =\dfrac{\sum \left (x_i-\bar x \right )^{2} }{n - 1}$

${\sum \left (x_i-\bar x \right )^{2} }$ = 453.333

n = 15

s² = 453. $\overline 3$ /(15 - 1) = 32.38

Part B

The measure of central tendency are the mean and the median because the size of the data (n = 15), is 75% of the population (N = 20) nd therefore the data is approximately normal and can be represented by the mean and the standard deviation

Step-by-step explanation:

4 0

3 years ago

The U.S. Bureau of Labor Statistics reports that of persons who usually work full-time, the average number of hours worked per w

alukav5142 [94]

Answer:

The standard deviation of number of hours worked per week for these workers is 3.91.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by

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

After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.

In this problem we have that:

The average number of hours worked per week is 43.4, so $\mu = 43.4$ .

Suppose 12% of these workers work more than 48 hours. Based on this percentage, what is the standard deviation of number of hours worked per week for these workers.

This means that the Z score of $X = 48$ has a pvalue of 0.88. This is Z between 1.17 and 1.18. So we use $Z = 1.175$ .