[FIXED] How can generate n numbers in the interval [a,b] that their summation doesn't exceed k in Python?

Issue

Let me I from the question in an easy way:

I usually have about k=3000€ per month. This month had n=26 working days (in July as you see in the following picture), and generally, I work something between [100,120]€ each day.

Note: k could be +/- x€ if needed, but it should be as minimum as possible.

what I tried to generate n numbers within [a,b] interval, but it should be very close to the k:

import numpy as np
#rng = np.random.default_rng(123)
#arr1 = rng.uniform(100, 120,26)
arr1 = np.random.randint(100,120,26)

#array([107, 115, 116, 105, 104, 110, 110, 107, 116, 110, 101, 112, 109,
#       111, 118, 102, 108, 113, 101, 112, 111, 116, 111, 109, 110, 107])

total = np.sum(arr1)
print(f'Sum of all the elements is {total}')
#Sum of all the elements is 2851

I don't have any clue to fulfil the condition. The summation of generated random numbers should be close to k [k, k+i] i=minimum e.g. [3000€, 3050€].

Edit1: I tried to compare the distribution quality of generated values offered by plotting/fitting offered solutions from @Murali & @btilly in the form of PDF as below:

import numpy as np
import scipy.stats as stats
import matplotlib.pyplot as plt

h = arr1
h.sort()
hmean = np.mean(h)
hstd = np.std(h)
pdf = stats.norm.pdf(h, hmean, hstd)
#plt.hist(arr1)
plt.plot(h, pdf,'-o',alpha=0.4) # including h here is crucial

So clearly one has a skew, but the other is the normal distribution.

Solution

You can use a Gaussian distribution for a given mean and standard deviation

mu = 3000/26

sigma = 5 ## allowed deviation from mean +- 5 from mean i.e [110.4,120.4]

arr1 = np.random.normal(mu, sigma, 26)

print(np.sum(arr1))
# 3011.268333226019

You can also play with other distributions and see which fits your purpose.

Answered By - Murali