On a statistics examination the mean was 78 and the standard deviation was 10. (a) Determine the standard scores of two students whose grades were 93 and 62, respectively, (b) Determine the grades of two students whose standard scores were 0.6 and 1.2, respectively.

Respuesta :

Answer:

a) Grade = 93, Standard score = 1.5

Grade = 62, Standard score = -1.6

b) Standard scores = 0.6, Grade = 84

Standard scores = 1.2, Grade = 90

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 78

Standard Deviation, σ = 10

Formula:

[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]

a) Grade = 93

[tex]\text{Standard score } = \displaystyle\frac{93-78}{10} = 1.5[/tex]

Grade = 62

[tex]\text{Standard score } = \displaystyle\frac{62-78}{10} = -1.6[/tex]

b) We have to find grades of student

Standard scores = 0.6

[tex]\displaystyle\frac{x-78}{10} = 0.6\\\\x = (0.6)(10)+78 \\x = 84[/tex]

Grade = 84

Standard scores = 1.2

[tex]\displaystyle\frac{x-78}{10} = 1.2\\\\x = (1.2)(10)+78 \\x = 90[/tex]

Grade = 90

The grade = 93, Standard score = 1.5,grade = 62,

Standard score = -1.6,Standard scores = 0.6, Grade = 84 and Standard scores = 1.2,

Grade = 90

We have given that,Mean( μ )= 78,Standard Deviation(σ) = 10

What is the formula Z score?

[tex]Z=\frac{x-\mu }{\sigma }[/tex]

a) Grade = 93

[tex]slandered score=\frac{93-78}{10} =1.5[/tex]

Grade = 62

[tex]slandered score=\frac{63-78}{10} =-1.6[/tex]

b) We have to find grades of student

Standard scores = 0.6

[tex]\frac{x-78}{10}=0.6[/tex]

[tex]x=(0.6)(10)+78[/tex]

Grade = 84

Standard scores = 1.2

[tex]\frac{x-78}{10}=1.2[/tex]

[tex]x=(1.2)(10)+78[/tex]

x=90

Grade = 90

Therefore we get the grade = 93, standard score = 1.5,grade = 62, Standard score = -1.6,Standard scores = 0.6, Grade = 84 and standard scores = 1.2, Grade = 90

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