Answer:
0.1946 is the probability that the September energy consumption level is between 1100 kWh and 1250 kWh.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 990 kWh
Standard Deviation, σ = 198 kWh
We are given that the distribution of energy consumption levels is a bell shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
P(September energy consumption level is between 1100 kWh and 1250 kWh)
[tex]P(1100 \leq x \leq 1250)\\\\ = P(\displaystyle\frac{1100 - 990}{198} \leq z \leq \displaystyle\frac{1250 -990}{198}) \\\\= P(0.5556 \leq z \leq 1.3131)\\\\= P(z \leq 1.3131) - P(z < 0.5556)\\\\= 0.9054- 0.7108= 0.1946[/tex]
0.1946 is the probability that the September energy consumption level is between 1100 kWh and 1250 kWh.
