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In an attempt to judge and monitor the quality of instruction, the administratio

ID: 3050547 • Letter: I

Question

In an attempt to judge and monitor the quality of instruction, the administration of Mega-Byte Academy devised an examination to test students on the basic concepts that all should have learned. Each year, a random sample of 10 graduating students is selected for the test. The average score is used to track the quality of the educational process. Test results for the past 10 years are shown in the table below:

                                                                  

                                                                  Student

Year

1

2

  3

4

5

6

7

8

9

10

Average

1

89

91

71

89

97

62

93

87

69

82

    83.0

2

94

90

76

88

79

93

86

87

92

63

    84.8

3

97

86

85

88

65

87

76

87

81

71

    82.3

4

94

85

56

77

89

72

71

61

92

97

    79.4

5

60

57

79

83

64

92

86

64

92

74

    75.1

6

85

88

77

69

54

90

97

72

64

60

    75.6

7

62

67

78

63

89

93

71

59

93

84

    75.9

8

67

81

90

55

71

71

86

98

51

90

    76.0

9

90

70

59

88

46

83

63

94

72

70

    73.5

10

60

57

92

87

70

61

56

58

63

70

    67.5

Use these data to estimate the center and standard deviation for this distribution. Then calculate the two-sigma control limits for the process average. What comments would you make to the administration of the Mega-Byte Academy?

a) The center of the distribution is centered as _______. The Standard deviation is __________. (enter your responses rounded to one decimal place).

Calculate two-sigma control limits for the process average.

b) The UCL equals _____ and the LCL equals _____. (enter your responses rounded to one decimal place).

Please show answers step-by-step.

                                                                  

                                                                  Student

Year

1

2

  3

4

5

6

7

8

9

10

Average

1

89

91

71

89

97

62

93

87

69

82

    83.0

2

94

90

76

88

79

93

86

87

92

63

    84.8

3

97

86

85

88

65

87

76

87

81

71

    82.3

4

94

85

56

77

89

72

71

61

92

97

    79.4

5

60

57

79

83

64

92

86

64

92

74

    75.1

6

85

88

77

69

54

90

97

72

64

60

    75.6

7

62

67

78

63

89

93

71

59

93

84

    75.9

8

67

81

90

55

71

71

86

98

51

90

    76.0

9

90

70

59

88

46

83

63

94

72

70

    73.5

10

60

57

92

87

70

61

56

58

63

70

    67.5

Explanation / Answer

a)

centre = 77.31

standard deviation = s = standard deviation of 100 numbers

b)

sd for mean = s/sqrt(10)

UCL = Xbar + z* s_xbar

LCL = Xbar - z* s_xbar

Xbar = 77.31, z = 2 , s_xbar = s/sqrt(10)

hence you just find s and plug the values

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