Given that the selected sample (i.e. 5 packages) is relatively small and we are not told that the distribution is normal. We apply t-distribution.
t-score =
[tex] \frac{132.6-50}{ \frac{150}{ \sqrt{5} } } =-0.2594
[/tex]
The p-value of the t score with degree of freedom of 4 is given by 0.4041.
Therefore, the probability that the average weight of the chocolate in the 5 packages is less than the advertised 132.6 grams is 0.4041.
