Question

The mean distance for tee shots on the 1999 men’s PGA tour was 270 yards with a standard deviation of 8 yards. Assuming that the 1999 tee-shot distances are normally distributed:show work please

a) find the percentage of such tee shots that went between 265 and 285 yards

b) find the percentage that went greater than 290 yards

Answer 1

solution

(A)P(265< x <285 ) = P[(265-270) /8 < (x - ) / < (285-270) / 8)]

= P(-0.625 < Z <1.875 )

= P(Z <1.875 ) - P(Z <-0.625 )

Using z table   

= 0.9696-0.266

= 0.7036

answer=70.36%

(B)P(x >290 ) = 1 - P(x<290 )

= 1 - P[(x -) / < (290 -270) /8 ]

= 1 - P(z < 2.5)

Using z table

= 1 - 0.9938

= 0.0062

answer= 0.62%