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d1i1m1o1n [39]
3 years ago
10

The monthly electrical utility bills of all customers for the Far East Power and Light Company are known to be normally distribu

ted with mean equal to $87.00 a month and standard deviation of $36.00. If a statistical sample of n = 100 customers is selected at random, what is the probability that the average bill for those sampled will exceed $75.00? g
Mathematics
1 answer:
omeli [17]3 years ago
3 0

Answer:  0.9996

Step-by-step explanation:

Given : The monthly electrical utility bills of all customers for the Far East Power and Light Company are known to be normally distributed with  a mean \mu=$\ 87.00

Standard deviation : \sigma=$\ 36.00

Sample size : n=100

Let X be the random variable that represents the electricity utility bill for a randomly selected month .

z-score : z=\dfrac{X-\mu}{\dfrac{\sigma}{\sqrt{n}}}

For X = $75.00

z=\dfrac{75.00-87.00}{\dfrac{36}{\sqrt{100}}}\approx-3.33

Now, the probability that the average bill for those sampled will exceed $75.00 will be :-

P(X>75)=P(z>-3.33)=1-P(z\leq-3.33)\\\\=1-  0.0004342=0.9995658\approx0.9996

Hence, the probability that the average bill for those sampled will exceed $75.00 =0.9996

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