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](https://stats.stackexchange.com/questions/180057) 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](https://en.wikipedia.org/wiki/Chebyshev%27s_inequality#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 $99.1\%.$

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

<!-- language: lang-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))