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?

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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arithmetic sequence: a sequnce in which the diffrence between any two consecutive terms is the same

step-by-step explanation:

7 weeks

step-by-step explanation:

convert each fraction to the least common multiple of 20, so 3/4 = 15/20 & 1/10 = 2/20, then divide 15 by 2 and you get a little over 7, so the food will last 7 weeks.