4x - 3 - 3x = 12 + 2х

Which box plot correctly displays the data set with a maximum of 48, a minimum of 27, a median of 36, an upper quartile of 45, a

Vitek1552 [10]

Answer:

Image 1

Step-by-step explanation:

To answer this question, we need to know how to use a box plot.

I've attached an image that demonstrates this well and can teach you what each part of the box plot means.

Specifically, the 2 farthest dots at the end are the minimum and maximum. The two sides where the box starts are the lower and upper quartiles. Finally, the line in the middle of the box is the median.

With this, we can analyze each box plot and determine which one is correct.

Box Plot 1: Everything is plotted correctly- Maximum and minimum plotted correctly, median plotted correctly, upper and lower quartile plotted correctly.

Box Plot 2: Everything plotted correctly EXCEPT maximum of 48 is represented at 46.

Box Plot 3: Everything plotted correctly EXCEPT the upper quartile of 45 is at 39.

Box Plot 4: Upper quartile represented at 39 and maximum represented at 46 are both incorrect.

4 0

3 years ago

According to survey data, the distribution of arm spans for males is approximately Normal with a mean of 71.4 inches and a stan

Gekata [30.6K]

Answer:

a) 89.97% of men have arm spans between 66 and 76 inches.

b) The z-score for this person's arm span is 5.68. 0% of males have an arm span at least as long as this person

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions 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:

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

The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Mean of 71.4 inches and a standard deviation of 3.1 inches.

This means that $\mu = 71.4, \sigma = 3.1$

a. What percentage of men have arm spans between 66 and 76 inches?

The proportion is the pvalue of Z when X = 76 subtracted by the pvalue of Z when X = 66. The percentage is the proportion multiplied by 100.

X = 76

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

$Z = \frac{76 - 71.4}{3.1}$

$Z = 1.48$

$Z = 1.48$ has a pvalue of 0.9306

X = 66

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

$Z = \frac{66 - 71.4}{3.1}$

$Z = -1.74$

$Z = -1.74$ has a pvalue of 0.0409

0.9306 - 0.0409 = 0.8997

0.8997*100% = 89.97%

89.97% of men have arm spans between 66 and 76 inches.

b. A particular professional basketball player has an arm span of almost 89 inches. Find the z-score for this person's arm span. What percentage of males have an arm span at least as long as this person?

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

$Z = \frac{89 - 71.4}{3.1}$