 Solution

The given calibration yields $M_V=-3.5{\log }_{10}50\simeq -5.946$. The distance modulus is $31.446$, giving $\boxed{d\simeq 19.5\mathrm{Mpc}}$ if extinction is negligible. Hubble's law predicts $v_H\simeq 1400$ km s${}^{-1}$, compared with 1150 km s${}^{-1}$ observed. The approximately $-250$ km s${}^{-1}$ difference can be a line-of-sight peculiar velocity of the cluster and observer, together with distance-calibration uncertainties. The local flow need not follow a perfectly smooth expansion law.