Mean delivery time = 50 minutes
Variance = 25 minutes
probability that a randomly selected parcel will take 60 minutes to deliver
Variance = 25 minutes
Therefore, Standard deviation "= \\sqrt{25} =5"
"Z score = \\frac{( Given \\, value - Mean)}{Standard \\,deviation}"
Z score for 60 "=\\frac{ (60 - 50)} {5} = 2"
"Z(2) =" 0.97725
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