part 1 of 5\nthe test scores from a history test are 88, 95, 92, 60, 86, 78, 95, 98, 92, 96, 70, 80, 89, and…

part 1 of 5\nthe test scores from a history test are 88, 95, 92, 60, 86, 78, 95, 98, 92, 96, 70, 80, 89, and 96.\na. find the mean and standard deviation of the test scores.\nb. find the five - number summary of the test scores.\nc. describe the type of distribution. explain.\nd. do you think the test was an easy test or a hard test for these students? explain.\na. the mean is .\n(round to the nearest hundredth as needed.)

part 1 of 5\nthe test scores from a history test are 88, 95, 92, 60, 86, 78, 95, 98, 92, 96, 70, 80, 89, and 96.\na. find the mean and standard deviation of the test scores.\nb. find the five - number summary of the test scores.\nc. describe the type of distribution. explain.\nd. do you think the test was an easy test or a hard test for these students? explain.\na. the mean is .\n(round to the nearest hundredth as needed.)

Answer

Explanation:

Step1: Calculate the sum of scores

The scores are 88, 95, 92, 60, 86, 78, 95, 98, 92, 96, 70, 80, 89, 96. Sum = (88 + 95+92 + 60+86+78+95+98+92+96+70+80+89+96) = 1225

Step2: Calculate the mean

There are (n = 14) scores. Mean (\bar{x}=\frac{\text{Sum}}{n}=\frac{1225}{14}\approx87.50)

Step3: Calculate the variance

First, find the deviation of each score from the mean ((x_i-\bar{x})), square it ((x_i - \bar{x})^2), sum these squared - deviations (\sum_{i = 1}^{n}(x_i-\bar{x})^2). ((88 - 87.5)^2=0.25), ((95 - 87.5)^2 = 56.25), ((92 - 87.5)^2=20.25), ((60 - 87.5)^2 = 756.25), ((86 - 87.5)^2=2.25), ((78 - 87.5)^2 = 90.25), ((95 - 87.5)^2=56.25), ((98 - 87.5)^2 = 110.25), ((92 - 87.5)^2=20.25), ((96 - 87.5)^2 = 72.25), ((70 - 87.5)^2 = 306.25), ((80 - 87.5)^2 = 56.25), ((89 - 87.5)^2=2.25), ((96 - 87.5)^2 = 72.25) (\sum_{i = 1}^{14}(x_i - \bar{x})^2=0.25+56.25+20.25+756.25+2.25+90.25+56.25+110.25+20.25+72.25+306.25+56.25+2.25+72.25 = 1629) Variance (s^2=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n - 1}=\frac{1629}{13}\approx125.31)

Step4: Calculate the standard deviation

Standard deviation (s=\sqrt{s^2}=\sqrt{125.31}\approx11.19)

Answer:

a. The mean is 87.50 and the standard deviation is approximately 11.19. b. First, order the data: 60, 70, 78, 80, 86, 88, 89, 92, 92, 95, 95, 96, 96, 98. The minimum is 60. The first - quartile (Q_1): The lower half of the data is 60, 70, 78, 80, 86, 88, 89. The median of the lower half is 80. The median (Q_2): Since (n = 14), the median is the average of the 7th and 8th ordered values, (\frac{89 + 92}{2}=90.5) The third - quartile (Q_3): The upper half of the data is 92, 95, 95, 96, 96, 98. The median of the upper half is 95. The maximum is 98. The five - number summary is 60, 80, 90.5, 95, 98. c. To describe the distribution: The inter - quartile range (IQR=Q_3 - Q_1=95 - 80 = 15). The lower fence (=Q_1-1.5\times IQR=80-1.5\times15 = 80 - 22.5=57.5) The upper fence (=Q_3 + 1.5\times IQR=95+1.5\times15=95 + 22.5 = 117.5) Since there are no outliers ((60>57.5) and (98<117.5)) and the mean (87.50) is less than the median (90.5), the distribution is left - skewed. d. The mean score is 87.50. Since the mean is relatively high, it seems that the test was an easy test for these students. However, the presence of a score as low as 60 indicates that not all students found it easy. Overall, considering the relatively high mean, it can be said that, on average, the students performed well, suggesting that the test was likely on the easier side for this group of students.