1)
Answer)
As the data is normally distributed we can use standard normal z table to estimate the answers
Z = (x-mean)/s.d
Given mean = 12837
S.d = 1500
P(13600<x<14600) = p(x<14600) - p(x<13600)
P(x<14600)
Z = (14600 - 12837)/1500 = 1.18
From z table, P(z<1.18) = 0.881
P(x<13600)
Z = (13600 - 12837)/1500 = 0.51
From z table, P(z<0.51) = 0.695
Required probability is 0.881 - 0.695 = 0.186
2)
Mean = 982
S.d = 180
P(x>1290)
Z = (1290 - 982) /180 = 1.71
From z table, P(z>1.71) = 0.0436
3)
Mean = 74
S.d = 3
P(x>72)
Z = (72-74)/3 = -0.67
From z table, p(z>-0.67) = 0.7486
Full-time Ph.D. students receive an average of $12,837 per year. If the average salaries are normally...
Full-time Ph.D. students receive an average of $12,837 per year. If the average salaries are normally distributed with a standard deviation of $1500, find the probability that the student makes more than $15,000. Group of answer choices A.0.0747 B.0.6051 C.0.0834 D.0.0982
Full-time Ph.D. students receive an average of $12,837 per year. If the average salaries are normally distributed with a standard deviation of $1500, find: B) the probability that a student makes between $13,000 and $14,000. (C) the interquartile range for the salaries. (D) the salaries belonging to the 10th and 90th percentiles. (Be sure you calculate both values!) Can you also please provide the equations needed? would be a big help thank you. Also, would this type of question be...
Full-time Ph.D. students receive an average of $12,837 per year. If the average salaries are normally distributed with a standard deviation of $1500, find: B) the probability that a student makes between $13,000 and $14,000. (C) the interquartile range for the salaries. (D) the salaries belonging to the 10th and 90th percentiles. (Be sure you calculate both values!) B I know is 0.2385 (correct answer) I'm having a hard time with C and D. Can you also please provide the...
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