Because (a) the number of events in any $T$ hour period with a rate of one event every $H$ hours follows a Poisson distribution with parameter $T/H$ and (b) the number of independent Poisson random variables of rates $\mu_1,\mu_2,\ldots,\mu_n$ is a Poisson random variable of rate $\mu=\mu_1+\mu_2+\cdots+\mu_n,$ any question about the probability distribution of the number of events is readily answered. Let's write $F$ for this distribution function and (therefore) $F^{-1}$ for its quantile function.

For instance,

• The expected number of events is $\mu.$

• The variance of the expectation is $\mu$ and so its standard deviation is $\sqrt{\mu}.$

• For any tolerance level $q$ between $0$ and $100%,$ choose nonnegative values $\alpha_l$ and $\alpha_u$ totaling $1-q.$ Then with at least probability $q$ the number of events will be between $F_{l} =F^{-1}(\alpha_l)$ and $F_{u}=F^{-1}(1-\alpha_u).$ The actual probability is $F(F_{u}) - F(F_{l}-1).$

Often one chooses a "symmetrical" interval in the sense that $\alpha_l=\alpha_u = (1-q)/2.$

To see how simple this is, let's use the data in the question as an example.

1. The time intervals are $100,80,130,50,400$ minutes and the rates are the reciprocals of $500,700,300,800,900$ hours, whence $$\mu = \frac{100}{500} + \frac{80}{700} + \cdots + \frac{400}{900} = 1.2545\ldots\,.$$

2. The standard deviation is $\sqrt{1.2545\ldots} \approx 1.12.$ You can use this immediately (using only mental arithmetic) to bound the chances of large numbers of events. For instance, Cantelli's Inequality implies the chance of $5$ or more events, which is $(5-1.2545)/1.12 \approx 3.3$ standard deviations beyond the mean, cannot exceed $1/(1+3.3^2) \approx 1/12 \approx 8\%.$

3. If, say, you wanted a $q=95\%$ symmetric interval in which the number of events is likely to fall, set $\alpha_l=\alpha_u=2.5\%$ and locate the $0.025$ and $0.975$ quantiles of the Poisson$(1.2545)$ distribution at $0$ and $4,$ respectively. The actual chance that the number of events will be between $0$ and $4$ inclusive is $99.1\%.$

To perform the calculations in the example I used the following commands in the statistical calculator R.

#
# Specify the problem.
#
intervals <- c(100, 80, 130, 50, 400)
averages <- c(500, 700, 300, 800, 900)
q <- 0.95
#
# Compute some properties of the distribution of the total number of events.
#
mu <- sum(intervals / averages)                    # Expected number
alpha.u <- alpha.l <- (1 - q)/2                    # Upper and lower tolerances
tl <- qpois(c(Lower=alpha.l, Upper=1-alpha.u), mu) # Limits
coverage <- diff(ppois(tl + c(-1,0), mu))          # Probability within limits 
names(coverage) <- "Coverage"
#
# Display the answers.
#
(tl) # The limits
(signif(c(Expected=mu, SD=sqrt(mu), `Nominal coverage`=q, coverage), 3))

Because (a) the number of events in any $T$ hour period with a rate of one event every $H$ hours follows a Poisson distribution with parameter $T/H$ and (b) the number of independent Poisson random variables of rates $\mu_1,\mu_2,\ldots,\mu_n$ in non-overlapping intervals is a Poisson random variable of rate $\mu=\mu_1+\mu_2+\cdots+\mu_n,$ any question about the probability distribution of the number of events is readily answered. Let's write $F$ for this distribution function and (therefore) $F^{-1}$ for its quantile function.

For instance,

• The expected number of events is $\mu.$

• The variance of the expectation is $\mu$ and so its standard deviation is $\sqrt{\mu}.$

• For any tolerance level $q$ between $0$ and $100%,$ choose nonnegative values $\alpha_l$ and $\alpha_u$ totaling $1-q.$ Then with at least probability $q$ the number of events will be between $F_{l} =F^{-1}(\alpha_l)$ and $F_{u}=F^{-1}(1-\alpha_u).$ The actual probability is $F(F_{u}) - F(F_{l}-1).$

Often one chooses a "symmetrical" interval in the sense that $\alpha_l=\alpha_u = (1-q)/2.$

To see how simple this is, let's use the data in the question as an example.

1. The time intervals are $100,80,130,50,400$ minutes and the rates are the reciprocals of $500,700,300,800,900$ hours, whence $$\mu = \frac{100}{500} + \frac{80}{700} + \cdots + \frac{400}{900} = 1.2545\ldots\,.$$

2. The standard deviation is $\sqrt{1.2545\ldots} \approx 1.12.$ You can use this immediately (using only mental arithmetic) to bound the chances of large numbers of events. For instance, Cantelli's Inequality implies the chance of $5$ or more events, which is $(5-1.2545)/1.12 \approx 3.3$ standard deviations beyond the mean, cannot exceed $1/(1+3.3^2) \approx 1/12 \approx 8\%.$

3. If, say, you wanted a $q=95\%$ symmetric interval in which the number of events is likely to fall, set $\alpha_l=\alpha_u=2.5\%$ and locate the $0.025$ and $0.975$ quantiles of the Poisson$(1.2545)$ distribution at $0$ and $4,$ respectively. The actual chance that the number of events will be between $0$ and $4$ inclusive is $99.1\%.$

To perform the calculations in the example I used the following commands in the statistical calculator R.

#
# Specify the problem.
#
intervals <- c(100, 80, 130, 50, 400)
averages <- c(500, 700, 300, 800, 900)
q <- 0.95
#
# Compute some properties of the distribution of the total number of events.
#
mu <- sum(intervals / averages)                    # Expected number
alpha.u <- alpha.l <- (1 - q)/2                    # Upper and lower tolerances
tl <- qpois(c(Lower=alpha.l, Upper=1-alpha.u), mu) # Limits
coverage <- diff(ppois(tl + c(-1,0), mu))          # Probability within limits 
names(coverage) <- "Coverage"
#
# Display the answers.
#
(tl) # The limits
(signif(c(Expected=mu, SD=sqrt(mu), `Nominal coverage`=q, coverage), 3))