Question

The original true mean is 31 with a standard deviation of 9. With the top 12% of grades being an A, the next 20% are Bs, the next 35% are Cs, the next 25% are Ds, and the rest (8%) should not pass. (a) Between what two scores would a student earn a B? (b) If the goal is to generate the same percentages for all grades as the original distribution, what would the curved average and standard deviation be if the curved range for B grades was set between 80 and 90?

Answer 1

a) score between top 12 to top 20+12=top 32%

This is a normal distribution question with

P(X < x) = 0.88

This implies that

P(Z < 1.1750) = 0.88

With the help of formula for z, we can say that

x = 41.5749

P(X < x) = 0.68

This implies that

P(Z < 0.4677) = 0.68

With the help of formula for z, we can say that

x = 35.2093

Between x = 35.21 and 41.57

d) Given in the question

P(X < x) = 0.9

This implies that

P(Z < 1.2816) = 0.9

With the help of formula for z, we can say that

x = 42.534

P(X < x) = 0.8

This implies that

P(Z < 0.8416) = 0.8

With the help of formula for z, we can say that

x = 38.57

Between 38.57 to 42.53

PS: you have to refer z score table to find the final probabilities.

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