Question

1) Two groups of second grade students were given a counting task. Group A was given objects and pictures to work with to help them count. Group B was just given picture representations, and took longer overall to complete the task. Each of the histograms below show the distribution of completion times, in minutes, for one of the two groups. Histogram Histogram II 12 Frequency Frequency 35 45 55 65 75 85 95 105115 35 45 55 65 75 85 95 105115 Completion Time Completion Time a) Of the two histograms shown, I and II, which is more likely to be the distrībution of completion times for the students in Group A? Justify your answer. b)Describe the shape of a histogram created from the data of the two groups of students combined. c) Consider the population of all students in second grade in this state who are given the same task. The distribution of the completion times has a shape similar to the combined histogram of students at the high school, with mean 70 minutes and standard deviation 26.5 minutes. For random samples of 50 students taken from the population, describe the sampling distribution of the sample mean completion time.

Answer 1

1.

a) In histogram I we can clearly see that the number of students who complete the task in less time is high. This is because their frequency is high. Moving along the graph we can see that the number of students who take time between 65 and 115 minutes is less.

In histogram II we can see that the frequency of students who complete their task in less time is low which means that the number of students who take less time to complete their task is less. The frequency of students who take more time between 95 and 115 is high.

In the problem it is given that the students in Group B take more time to complete the task and students in Group A take less time to complete the task.

As the students in Histogram I take less time to complete the task therefore distribution of Histogram 1 is group A

b) A combined histogram is made with both histograms and the distribution is not uniform or normal or skewed. It can be seen in the images below.

tegy and b) In Histogram I and II, the time and is listed below by taking values from the chistogram Histogram Histogram Time) Frequency 1 (Time) Scarls Scores Frequency 35 35 45 2 45 12 55 11 55 65 3 65 75 2 75 2 85 2 85 3 95 105 95 105 و تا 115. 115. 9 Total. 42. 42 Now add to combine the histogram, the frequencies corresponding same times .

act Time Frequen гем. 8 35 8+1=9 12+2=14 45 litl=12 55 65 3+1=4 75 2+2=4 85 2+3=5 95 2tlo = 12 105 1+13=14 115 (+9 Total 84 Now we will plot this listograun.

14 14 2 Frequenly 10 lo +9 8 6 44 2 we can say that not 35 45 55 65 45 85.95. 105 115. Time . Upon plotting that it is normally distributed shape. The frequency is high in the beginning and end but low in the middle.

c) Since we are taking the means of 50 samples and then plotting it.

It is given that the distribution is similar to the distribution in b part. This means that the sampling distribution is not normal or skewed. The sample means are very different from the mean of sampling distribution. This can be seen from the fact that the mean of new distribution is 70 and standard deviation is 26.5 . This high amount of standard deviation is possible in the distribution similar to the combined distribution in b part.

2.

a) Looking at the histogram we can say that the distribution is a normal distribution which is slightly right skewed. Note that we are also considering the outlier value of 7 to 8 in this distribution.

b) Inter quartie range is calculated by subtracting Q1 from Q3.

Therefore IQR = Q3 - Q1 = 3.8 - 2.6 = 1.2

To check whether we have any outliers we will subtact 1.5 IQR from lower quartile and add 1.5 IQR to upper quartile.

Lower limit = Q1 - 1.5 IQR = 2.6 - (1.5)*(1.2) = 2.6 - 1.8 = 0.8

Since the minimum value is 1.4 therefore there is no outlier in lower range.

Upper limit = Q3 + 1.5 IQR = 3.8 + (1.5)*(1.2) = 3.8 + 1.8 = 5.6

As the maximum value is 7.5 , therefore it is an outlier as it crosses the range.

So we can say that this method is good to check the outliers.