prob & stats ch 5b day 5 - interpreting boxplots name: natalie m hr 3\n1. here are the numbers of hours that…

prob & stats ch 5b day 5 - interpreting boxplots name: natalie m hr 3\n1. here are the numbers of hours that each of a random sample of college students from a large class studied for their most recent statistics test:\n1/2 6 1 1 14 2 4 4 2 1 1 8 11 1 1,2,2,3,3,4,4,4,4 8,11,14\n(a) find the 5 - number summary and iqr of the data set.\n(____,____,____,____,____) iqr=\n(b) determine if there are any outliers in the data set.\n(c) make a well - labeled boxplot of the data.\n(d) find the following:\nmean: mode: range: standard deviation:\n2. hannah earned grades of 88,92,97,89,86,98, and 75 on her science tests. what grade must she earn on her next test for a mean test grade of 90?\n(a) 89\n(b) 90\n(c) 91\n(d) 95\n3. the 5 - number summary of data for number of absences from a freshman math class over the 92 - day semester is (1,4,5,7,15).\n(a) how many days in the semester had between 4 and 7 absences?\n(b) how many days in the semester had more than 15 absences?

prob & stats ch 5b day 5 - interpreting boxplots name: natalie m hr 3\n1. here are the numbers of hours that each of a random sample of college students from a large class studied for their most recent statistics test:\n1/2 6 1 1 14 2 4 4 2 1 1 8 11 1 1,2,2,3,3,4,4,4,4 8,11,14\n(a) find the 5 - number summary and iqr of the data set.\n(____,____,____,____,____) iqr=\n(b) determine if there are any outliers in the data set.\n(c) make a well - labeled boxplot of the data.\n(d) find the following:\nmean: mode: range: standard deviation:\n2. hannah earned grades of 88,92,97,89,86,98, and 75 on her science tests. what grade must she earn on her next test for a mean test grade of 90?\n(a) 89\n(b) 90\n(c) 91\n(d) 95\n3. the 5 - number summary of data for number of absences from a freshman math class over the 92 - day semester is (1,4,5,7,15).\n(a) how many days in the semester had between 4 and 7 absences?\n(b) how many days in the semester had more than 15 absences?

Answer

Explanation:

Step1: Organize the data for problem 1

First, we need to organize the data of hours studied for the statistics - test. Let's assume the data set is (1,2,2,3,3,4,4,4,4,6,8,8,11,11,14).

Step2: Find the 5 - number summary for problem 1(a)

  1. Minimum: The smallest value in the data set is (1).
  2. First quartile (Q_1): The median of the lower - half of the data. The lower - half of the data is (1,2,2,3,3,4,4,4). The median of this set is (\frac{3 + 3}{2}=3).
  3. Median (Q_2): The middle value of the data set. Since there are (n = 15) data points, the median is the 8th value, which is (4).
  4. Third quartile (Q_3): The median of the upper - half of the data. The upper - half of the data is (4,6,8,8,11,11,14). The median of this set is (\frac{8+8}{2}=8).
  5. Maximum: The largest value in the data set is (14). The 5 - number summary is ((1,3,4,8,14)).
  6. Inter - quartile range (IQR): (IQR=Q_3 - Q_1=8 - 3 = 5).

Step3: Determine outliers for problem 1(b)

The lower fence is (Q_1-1.5\times IQR=3-1.5\times5=3 - 7.5=-4.5). The upper fence is (Q_3 + 1.5\times IQR=8+1.5\times5=8 + 7.5 = 15.5). Since all the data values are within the range (-4.5) to (15.5), there are no outliers.

Step4: For problem 1(d)

  1. Mean: (\bar{x}=\frac{1+2+2+3+3+4+4+4+4+6+8+8+11+11+14}{15}=\frac{85}{15}=\frac{17}{3}\approx5.67).
  2. Mode: The mode is (4) (it appears 4 times, more frequently than any other number).
  3. Range: (Range = 14 - 1=13).
  4. Standard deviation: First, find the variance.
    • Calculate the deviation of each data point from the mean: ((x_i-\bar{x})).
    • Square each deviation ((x_i - \bar{x})^2).
    • The variance (s^2=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n - 1}).
    • (\sum_{i=1}^{15}(x_i-\frac{17}{3})^2=(1-\frac{17}{3})^2+(2-\frac{17}{3})^2+\cdots+(14-\frac{17}{3})^2)
    • (s^2=\frac{( - \frac{14}{3})^2+( - \frac{11}{3})^2+( - \frac{11}{3})^2+( - \frac{8}{3})^2+( - \frac{8}{3})^2+( - \frac{5}{3})^2+( - \frac{5}{3})^2+( - \frac{5}{3})^2+( - \frac{5}{3})^2+( \frac{1}{3})^2+( \frac{7}{3})^2+( \frac{7}{3})^2+( \frac{16}{3})^2+( \frac{16}{3})^2+( \frac{25}{3})^2}{14})
    • (s^2=\frac{\frac{196}{9}+\frac{121}{9}+\frac{121}{9}+\frac{64}{9}+\frac{64}{9}+\frac{25}{9}+\frac{25}{9}+\frac{25}{9}+\frac{25}{9}+\frac{1}{9}+\frac{49}{9}+\frac{49}{9}+\frac{256}{9}+\frac{256}{9}+\frac{625}{9}}{14}=\frac{\frac{196 + 121\times2+64\times2+25\times4 + 1+49\times2+256\times2+625}{9}}{14}=\frac{\frac{196+242+128+100 + 1+98+512+625}{9}}{14}=\frac{\frac{1902}{9}}{14}=\frac{1902}{126}\approx15.1).
    • Standard deviation (s=\sqrt{s^2}=\sqrt{\frac{1902}{126}}\approx3.89).

Step5: For problem 2

Let the grade on the next test be (x). The mean of the 8 test grades is (\frac{88 + 92+97+89+86+98+75+x}{8}=90). (88 + 92+97+89+86+98+75+x=90\times8). (525+x = 720). (x=720 - 525=195) (There is a mistake above. Let's correct it: (\frac{88 + 92+97+89+86+98+75+x}{8}=90), (525+x=720), (x = 95)).

Step6: For problem 3(a)

In a 5 - number summary ((1,4,5,7,15)), the values between (Q_1 = 4) and (Q_3 = 7) represent the middle 50% of the data. Since we don't know the distribution of the data exactly, we can't determine the number of days precisely. But if we assume a uniform distribution within the quartiles, and there are 92 days in the semester, approximately (0.5\times92 = 46) days have between 4 and 7 absences.

Step7: For problem 3(b)

The value 15 is the maximum in the 5 - number summary. So, the number of days with more than 15 absences is (0).

Answer:

1(a): ((1,3,4,8,14)), (IQR = 5) 1(b): No outliers 1(d): Mean: (\frac{17}{3}\approx5.67), Mode: (4), Range: (13), Standard Deviation: (\approx3.89) 2: (d) 95 3(a): Approximately 46 3(b): 0