Can anyone help me answer this<br>

A manager recorded how many hours each of 11 cashiers worked in a week. Which interval contains the median number of hours worke

Lana71 [14]

Answer:

Step-by-step explanation:

Hello!

In descriptive statistics, the median is a measurement of center position. It separates the sample in halves. (bottom 50% from upper 50%)

The manager took a sample of 11 cashiers and recorded the worked hours in a week

To identify the median you have to calculate it's position:

The position is calculated as

PosMe= (n+1)/2= (11+1)/2= 6 cashier

This means that the Median of hours worked is a week observed corresponds to the 6th cashier (after ordering the data from lowest to highest)

If you observe the histogram, the x-axis corresponds to worked hours and the y-axis shows the number of workers.

The workers were already arranged from least hours worked to most hours worked.

The first interval corresponds to 0-10 worked hours and two workers recorded these working times.

The second interval corresponds to 10-20 worked hours and three workers recorded these working times.

In the third interval of 20-30 hours, no workers recorded these working times.

The fourth interval corresponds to 30 - 40 worked hours and 5 workers recorded these working times.

The fifth interval corresponds to 40 - 50 worked hours and only one worker recorded these working hours.

The first five workers were recorded in the first two intervals and the sixth worker was recorded in the fourth interval. So the interval that contains the median number of hours worked is "30 - 40 hours"

I hope this helps!

5 0

3 years ago

Which is NOT a true statement?

rusak2 [61]

Answer:

option 4

Step-by-step explanation:

|a| represents Absolute value, it is the distance from zero that a number is on the number line. And we know distance cannot be negative.

So,

$|a| = a$

$|-a| = a\\\\| \frac{-a}{b}| = \frac{a}{b}\\\\|\frac{a}{-b}| = \frac{a}{b}$

6 0

3 years ago

Read 2 more answers

What’s the sum of interior angles of a 45-gon?

Mashutka [201]

Answer:

Step-by-step explanation:

As each exterior angle is 45o , number of angles or sides of the polygon is 360o45o=8 . Further as each exterior angle is 45o , each interior angle is 180o−45o=135o .

7 0

3 years ago

Find the length of side BC.<br> Give your answer to 1 decimal place.

Alenkasestr [34]

Use

Angle sin A = AB/BC

AND FIND IT OUT

PLEASE FOLLOW ME

5 0

3 years ago

Read 2 more answers

The overhead reach distances of adult females are normally distributed with a mean of 197.5 cm197.5 cm and a standard deviation

fiasKO [112]

Answer:

a) 5.37% probability that an individual distance is greater than 210.9 cm

b) 75.80% probability that the mean for 15 randomly selected distances is greater than 196.00 cm.

c) Because the underlying distribution is normal. We only have to verify the sample size if the underlying population is not normal.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are 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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

$\mu = 197.5, \sigma = 8.3$

a. Find the probability that an individual distance is greater than 210.9 cm

This is 1 subtracted by the pvalue of Z when X = 210.9. So

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

$Z = \frac{210.9 - 197.5}{8.3}$

$Z = 1.61$

$Z = 1.61$ has a pvalue of 0.9463.

1 - 0.9463 = 0.0537

5.37% probability that an individual distance is greater than 210.9 cm.

b. Find the probability that the mean for 15 randomly selected distances is greater than 196.00 cm.

Now $n = 15, s = \frac{8.3}{\sqrt{15}} = 2.14$

This probability is 1 subtracted by the pvalue of Z when X = 196. Then

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

By the Central Limit Theorem

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

$Z = \frac{196 - 197.5}{2.14}$

$Z = -0.7$

$Z = -0.7$ has a pvalue of 0.2420.

1 - 0.2420 = 0.7580

75.80% probability that the mean for 15 randomly selected distances is greater than 196.00 cm.

c. Why can the normal distribution be used in part (b), even though the sample size does not exceed 30?

The underlying distribution(overhead reach distances of adult females) is normal, which means that the sample size requirement(being at least 30) does not apply.

5 0

3 years ago

Can anyone help me answer this​

Can anyone help me answer this