 Solution

An open ion channel carries the electric currentThe macroscopic mean is $I=N{P}_{o}i$, so $I/i\simeq 204$ is the mean number open, not the total number present. Reading $P_o(-20\mathrm{mV})\simeq 0.40$ from the original graph givesThe graph reading is approximate, so a precision of hundreds is appropriate. Only a few hundred active ion channels generate a nanoampere-scale cellular electric current; molecular discreteness and stochastic gating therefore matter. For independent identical ion channels, the open-count standard deviation is $\sqrt{N{P}_o(1-P_o)}\simeq 11$, corresponding to a coefficient of variation in electric current of about $5.4\%$. This is an instantaneous ideal independent-gating estimate, before temporal filtering and other noise sources, and requires the single-channel recording $P_o$ to apply to the cellular recording conditions.