The average score of all golfers for a particular course has a mean of 73 and a standard deviation of 5. Suppose 100 golfers played the course today. Find the probability that the average score of the 100 golfers exceeded 74. Round to four decimal places.
Solution :
Given that,
mean = = 73
standard deviation = =5
n=100
= =73
= / n = 5/ 100 = 0.5
P( > 74) = 1 - P( < 74)
= 1 - P[( - ) / < (74-73) /0.5 ]
= 1 - P(z < 2)
Using z table
= 1 - 0.9772
probability= 0.0228
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