A bus company wants to use the data to show how long completing a bus route takes so that it can create a bus schedule.

Mathematics

1 answer:

Sergeu [11.5K]3 years ago

6 0

Bus company wants to use data to show how long a bus takes to complete the route:

The data set of the time taken to complete the route is given as follows:

31 31 31 33 31 32 32 60 31 32 33

Median: Median is the mid value of the data set when the data is arranged in the order from lowest to greatest. Mid value can help us create a bus schedule. Here in our case mean of the data set is 34.17 and there is another element in the data set that is 60 which is way beyond the average value.

Mean: Mean or the average is obtained by summing up all the values and dividing that by number of values in the data set. It can tell us how much time does the bus take on an average to complete the route and it can help up to create a bus schedule but in this case there are values which are way beyond the mean, therefore it is a little hard to consider mean to design a bus schedule.

Minimum: Minimum of a data set can not help us to create a bus schedule.

Range: Range is defined by the difference between the maximum and the minimum value of a data set, it can not help us to figure out how much time on an average does a bus take to complete its route.

So, in this case median can help us create a bus schedule.

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How many weeks of data must be randomly sampled to estimate the mean weekly sales of a new line of athletic footwear? We want 98

Irina18 [472]

Answer:

$n=(\frac{2.33(1400)}{400})^2 =66.50 \approx 67$

So the answer for this case would be n=67 rounded up

Step-by-step explanation:

Information given

$\bar X$ represent the sample mean for the sample

$\mu$ population mean

$\sigma=1400$ represent the sample standard deviation

n represent the sample size

Solution to the problem

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (a)

And on this case we have that ME =400 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (b)

The critical value for 98% of confidence interval now can be founded using the normal distribution. And the critical value would be $z_{\alpha/2}=2.33$ , replacing into formula (b) we got: