More data needed. SD?
Can assume mean = (9.5-6.5)/2?
Z = (score-mean)/SEm
SEm = SD/√n
Look in the back of your statistics textbook for a table called something like “area under normal distribution” to find the proportion/probability of the Z score.
What is the probability that simple random sample of 50 staff members provides a sample mean leave of 6.5 to 9.5 weeks?
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