Answer:
0.5252 is the required probability.
Step-by-step explanation:
The attached image shows the missing information of the question.
We are given the following information in the question:
Mean, μ = 20 minutes
Sample standard Deviation, s = 15.81 minutes
Sample size, n = 29
We are given that the distribution of delivery time is a bell shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{s}[/tex]
P(delivery time is atmost 21 minutes)
[tex]P( x \leq 21) = P( z \leq \displaystyle\frac{21 - 20}{15.81}) = P(z \leq 0.0632)[/tex]
Calculation the value from standard normal z table, we have,
[tex]P(x \leq 21) =0.5252 = 52.52\%[/tex]
o.5252 is the probability that the average delivery time of the selected delivered at most 21 minutes.
