Question

According to the International Atomic Energy Agency, per capita electricity consumption in Canada was 16,939 kWh in 2002. A random sample of 51 households was monitored for one year to determine electricity usage. If the population standard deviation of annual usage is 3,500 kWh, what is the probability that the sample mean will be each of the following? a.More than 18,000 kWh b.More than 17,500 kWh c.Between 17,000 kWh and 18,000 kWh d.Less than 16,000 kWh e.Less than 15,000 kWh

Answer 1

Given,

= 16,939 kWh

= 3,500 kWh

Sample of size, n = 51is drawn.

According to Central Limit Theorem, the sampling distribution of mean will be normal with,

Mean, = = 16,939 kWh

Standard deviation, =

=

= 490.1

P( < A) = P(Z < (A - )/)

a) P(more than 18,000 kWh) = P( > 18,000)

= 1 - P( < 18,000)

= 1 - P(Z < (18,000 - 16,939)/490.1)

= 1 - P(Z < 2.16)

= 1 - 0.9846

= 0.0154

b) P(more than 17,500) = 1 - P( < 17,500)

= 1 - P(Z < (17,500 - 16,939)/490.1)

= 1 - P(Z < 1.145)

= 1 - 0.8739

= 0.1261

c) P(between 17,000 kWh and 18,000 kWh) = P( < 18,000) - P( < 17,000)

= 0.8739 - P(Z < (17,000 - 16,939)/490.1)

= 0.8739 - P(Z < 0.125)

= 0.8739 - 0.5497

= 0.3242

d) P( < 16,000 kWh) = P(Z < (16,000 - 16,939)/490.1)

= P(Z < -1.916)

= 0.0277

e) P( < 15,000 kWh) = P(Z < (15,000 - 16,939)/490.1)

= P(Z < -3.956)

= 0