Question

Statistics!

Dr. Ped is a foot doctor in the town of Hoofit, where the practice of taking a mid-day walk has really caught on. He talked to several of his patients and was happy to discover that they were engaging in this healthful activity. He wanted to learn more about how much time his patients were typically spending on their walks (longer walks can be rough on the feet if appropriate footwear and walking technique are not used), so he decided to collect some data. He randomly selected 6 of his walking patients and asked them to record their walking time on a given Monday.

a.) Using the data below, please calculate the mean and standard deviation for this sample (1 point).

Joan=39,       Pat=48,       Sam=51,       Teri=58,       Lori=62,       Ann=69

b.) Based on the sample data above, what percent of the population (all his patients) likely have a walking time between Pat and Lori (1 point)?

Answer 1

Data given

Patient	Waiting time
Joan	39
Pat	48
Sam	51
Teri	58
Lori	62
Ann	69

(a)

The mean of a data set is the sum of the terms divided by the total number of terms. Using math notation we have:

Mean=Sum of terms / Number of terms

Sum of terms = 39 + 48 + 51 + 58 + 62 + 69 = 327

Number of terms = 6

Mean= 327 / 6

=109/ 2=54.5

Create the following table.

data	data-mean	(data - mean)2
39	-15.5	240.25
48	-6.5	42.25
51	-3.5	12.25
58	3.5	12.25
62	7.5	56.25
69	14.5	210.25

Find the sum of numbers in the last column to get.

(b) Walk time for Pat = 48 and Lori = 62

We need to find the percentage of population like to have a walking time between 48 and 62.  

Here since we do not have population data we are assuming that sample mean and sample std deviation are for the population and the data is normally distributed.

The probability that 48<X<62 is equal to the blue area under the curve.

Assuming  μ=54.5 and σ=10.7098 we have:

P ( 48< X < 62 )= P ( 48−54.5 < X−μ < 62−54.5 )= P ( (48−54.5)/ 10.7098 < (X−μ)/ σ < (62−54.5)/ 10.7098)

Z=(x−μ) / σ =  (48−54.5)/ 10.7098 =−0.61 and (62−54.5)/ 10.7098 =0.7

P ( 48< X <62 )=P ( −0.61< Z <0.7 )

Use the standard normal table to conclude that:

P ( −0.61< Z < 0.7 )=0.4871

Hence 48.71% of the population have a likely walking time between Pat and Lori based on the sample data.