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Answer:
Roughly speaking, in a normal distribution, a score that is 1 s.d. above the mean is equivalent to the 84th percentile. ... In other words, roughly 95 percent of students are within two standard deviations – positive or negative – of the average
Based on the given conditions, formulate:
Rearrange unknown terms to the left side of the equation:
Combine like terms:
Calculate the sum or difference:
Divide both sides of the equation by the coefficient of variable:
Answer: k=41/3
Hello : here is a solution
Still need more information..like a picture
