Refer to the table of body temperatures (degrees Fahrenheit). Is there some meaningful way in which each body temperature

A business owner collected data on the people working for her company. She determined the average number of minutes each employe

Naddika [18.5K]

Answer:

The comparison using median and IQR is best because one of the graphs is not symmetrical.

Step-by-step explanation:

The following information is missing

A box plot titled Number of Minutes Women Spend on Breaks. The number line goes from 30 to 70. The whiskers range from 30 to 54, and the box ranges from 34 to 50. A line divides the box at 48.

A box plot titled Number of Minutes Men Spend on Breaks. The number line goes from 30 to 70. The whiskers range from 30 to 68, and the box ranges from 36 to 60. A line divides the box at 48.

The business owner uses the median and IQR to determine the center and variability of the data sets. Which best describes the comparison?

The comparison would be more accurate using the mean and MAD because one of the graphs is symmetric.

The comparison would be more accurate using the mean and MAD because the median of both data sets is the same.

The comparison using median and IQR is best because one of the graphs is not symmetrical.

The comparison using median and IQR is best because the median is greater than the IQR for both data sets.

Mean and MAD are useful for comparison when both data sets are symmetrical.

In the women box plot the Q1 is at 34, the median is at 48, and the Q3 is at 50, so it is not symmetrical (the difference between the median and the Q1, and the Q3 and the median is not the same)

8 0

3 years ago

3. Last year, the numbers of skateboards produced per day at a certain factory were normally distributed with a mean of 20,500 s

Oxana [17]

Let x be a random variable representing the number of skateboards produced
a.) P(x ≤ 20,555) = P(z ≤ (20,555 - 20,500)/55) = P(z ≤ 1) = 0.84134 = 84.1%

b.) P(x ≥ 20,610) = P(z ≥ (20,610 - 20,500)/55) = P(z ≥ 2) = 1 - P(z < 2) = 1 - 0.97725 = 0.02275 = 2.3%

c.) P(x ≤ 20,445) = P(z ≤ (20,445 - 20,500)/55) = P(z ≤ -1) = 1 - P(z ≤ 1) = 1 - 0.84134 = 0.15866 = 15.9%

4 0

2 years ago

A teacher places n seats to form the back row of a classroom layout. Each successive row contains two fewer seats than the prece

Alex_Xolod [135]

Answer:

The number of seat when n is odd $S_n=\frac{n^2+2n+1}{4}$