The grades on a language midterm at Oak Academy are roughly symmetric with = 67 and 0 = 2.5. Ishaan scored 70 on the exam. Find the z-score for Ishaan's exam grade. Round to the nearest tenth.

Answers

Answer 1

Answer:  1.2

Step-by-step explanation:

Let X denotes the grades on a  language midterm at Oak Academy.

As per given ,

X follows normal distribution ( because it has symmetric graph)

[tex]\mu=67,\ \ \ \ \sigma= 2.5[/tex]

Formula for z : [tex]z=\dfrac{X-\mu}{\sigma}[/tex]

For X = 70, we have

[tex]z=\dfrac{70-67}{2.5}=\dfrac{3}{2.5}=1.2[/tex]

Hence, the z-score for Ishaan's exam grade = 1.2

Answer 2

z-scores are the corresponding standard normal variate's values for normal variates' values. The z-score for Ishaan's grade is 1.2

How to get the z scores?

If we've got a normal distribution, then we can convert it to standard normal distribution and its values will give us the z score.

If we have

[tex]X \sim N(\mu, \sigma)[/tex]

(X is following normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] )

then it can be converted to standard normal distribution as

[tex]Z = \dfrac{X - \mu}{\sigma}, \\\\Z \sim N(0,1)[/tex]

(Know the fact that in continuous distribution, probability of a single point is 0, so we can write

[tex]P(Z \leq z) = P(Z < z) )[/tex]

Also, know that if we look for Z = z in z tables, the p-value we get is

[tex]P(Z \leq z) = \rm p \: value[/tex]

For the given case, if we suppose the random variable X's values as scores of the grades on given language midterm at Oak Academy, then, as it is symmetric around value 67, with σ = 2.5, thus,

[tex]X = N(\mu = 67, \sigma = 2.5)[/tex]

Its given that Ishaan scored X = 70

The standard normal variate conversion of X is given by:

[tex]Z = \dfrac{X- \mu}{\sigma} = \dfrac{X - 67}{2.5}[/tex]

Thus, z-scores for Ishaan's score is: [tex]Z = \dfrac{70 - 67}{2.5} = \dfrac{3}{2.5} = 1.2[/tex]

Learn more about standard normal distribution here:

https://brainly.com/question/10984889


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Answers

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Answer:

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