The time required for Jim to commute to Lexington is well approximately by a nor

Question

The time required for Jim to commute to Lexington is well approximately by a normal distribution with a mean of 75 mins and a standard deviation of 6 mins. Suppose Jim always plans on 90 mins for his commute to a meeting in Lexington. I would like to know how often Jim is late.

A. The z-score based on the given information is calculated to be _ when rounded to two digits after the decimal point.

B. If Jim plans on 90 mins for his commute to Lexington, then the proportion of time he will be late is _. Give the final answer rounded to three digits after the decimal point and write as a proportion, not as a % value.

Explanation / Answer

A) as we know Z score =(X-mean)/std deviaiton

therefore  z-score based on the given information =(90-75)/6 =2.5

B) proportion of time he will be late is =P(Z>2.5) =0.0062 .

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