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The scores received by students completing a geology exam are normally distributed. If the average score is 122 and the standard deviation is 35, what percentage of students scored between 52 and 122?

Respuesta :

LammettHash
Notice that [tex]52=122-70=122-2\times35[/tex], so the probability is equivalent to

[tex]\mathbb P(-2<Z<0)[/tex]

where [tex]Z[/tex] follows the standard normal distribution. This is half the proportion of the students that fall within two standard deviations of the mean:

[tex]\mathbb P(-2<Z<0)=\dfrac12\mathbb P(|Z|<2)[/tex]

Recall the empirical rule, which states that approximately 95% of a normal distribution falls within two standard deviations of the mean. This means that

[tex]\mathbb P(-2<Z<0)\approx\dfrac12\times0.95=0.475=47.5\%[/tex]

of students scored between 52 and 122.
fichoh

The percentage of students who scored between 52 and 122 is 47.73%

Given that :

Mean score, μ = 122

Standard deviation, σ = 35

The percentage of student who scored between 52 and 122 can be defined as :

P(z < Z) - P(z < Z)

Where, Z = (x - μ) / σ

Therefore, percentage of student who scored between 52 and 122 equals :

For x = 52 ;

P(z < Z) = P[z < (52 - 122) / 35)] = P[z < - 2]

For x = 122 ;

P(z < Z) = P[z < (122 - 122) / 35)] = P[z < 0]

Hence ;

P(z < 0) - P(z < - 2)

Using the normal distribution table or P value calculators ;

P(z < 0) = 0.5

P(z < - 2) = 0.02275

Therefore ;

P(z < 0) - P(z < - 2) = [0.5 - 0.02275] = 0.47725

The percentage of students who scored between 52 and 122 is (0.47725 × 100%) = 47.725% = 47.73%

Learn more : https://brainly.com/question/14971859

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