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Answer :
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%
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Rewritten by : Barada
Notice that [tex]52=122-70=122-2\times35[/tex], so the probability is equivalent to
[tex]\mathbb P(-2
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
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
of students scored between 52 and 122.
[tex]\mathbb P(-2
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
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
of students scored between 52 and 122.
