Bailee Ortiz

2022-05-03

Trying to find the probability $b/w$ two scores. I know mapping to $z$-score is:

${z}_{11}=(11-10)/1.5=.67$

and

${z}_{14}=(14-10)/1.5=2.67$

So,

$\overline{x}=10,\text{}\sigma =\mathrm{1.5.}$

Would I simple subtract the two from each other? and so on the real number line it would range from 0 to 100 and that is how the distribution would be set up. So, 2.67−2=2?

On the z-table 2 is given a value of 0.9772.

My teacher likes to give things as either a proportion or a percentage so how might one be able to address the two. Like if this question was asking for the proportion of scores that fall between 11 and 14 or if the question said what's the percentage of scores that fall between 11 and 14. How might be able to address both scenarios?

Just confused on the ideas of proportion, probability, and percentage and how they are relate.

${z}_{11}=(11-10)/1.5=.67$

and

${z}_{14}=(14-10)/1.5=2.67$

So,

$\overline{x}=10,\text{}\sigma =\mathrm{1.5.}$

Would I simple subtract the two from each other? and so on the real number line it would range from 0 to 100 and that is how the distribution would be set up. So, 2.67−2=2?

On the z-table 2 is given a value of 0.9772.

My teacher likes to give things as either a proportion or a percentage so how might one be able to address the two. Like if this question was asking for the proportion of scores that fall between 11 and 14 or if the question said what's the percentage of scores that fall between 11 and 14. How might be able to address both scenarios?

Just confused on the ideas of proportion, probability, and percentage and how they are relate.

Addison Zamora

Beginner2022-05-04Added 10 answers

Hint:

$P(11<X<14)=P({z}_{11}<\frac{X-\mu}{\sigma}<{z}_{14})=P(Z<{z}_{14})-P(Z\le {z}_{11})$

$P(11<X<14)=P({z}_{11}<\frac{X-\mu}{\sigma}<{z}_{14})=P(Z<{z}_{14})-P(Z\le {z}_{11})$

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