Mathematics, 30.03.2020 17:02 King1Gates

In one region, the September energy consumption levels for single-family homes are found to be normally distributed with a mean of 1050 kWh and a standard deviation of 218 kWh. If 50 different homes are randomly selected, find the probability that their mean energy consumption level for September is greater than 1075 kWh.

Answers: 3

Show answers

Answers

Answer from: momoney9957

The required probability is 0.68793.

Step-by-step explanation:

Since we have given that

Mean = 1050 kWh

Standard deviation = 218 kWh

n = 50

$P(X1065)=P(z\dfrac{1065-1050}{\dfrac{218}{\sqrt{50}}}\\\\P(X1065)=P(z\dfrac{15}{30.83})\\\\P(X1065)=P(z0.486)=0.68793$

Hence, the required probability is 0.68793.

Answer from: clonetrooper099

The probability that their mean energy consumption level for September is greater than 1075 kWh is 0.20897.

Step-by-step explanation:

We are given that in one region, the September energy consumption levels for single-family homes are found to be normally distributed with a mean of 1050 kWh and a standard deviation of 218 kWh.

Also, 50 different homes are randomly selected.

Let $\bar X$ = sample mean energy consumption level

The z-score probability distribution for sample mean is given by ;

Z = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean energy consumption levels = 1050 kWh

$\sigma$ = standard deviation = 218 kWh

n = sample of homes = 50

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

Now, the probability that their mean energy consumption level for September is greater than 1075 kWh is given by = P( $\bar X$ > 1075 kWh)

P( $\bar X$ > 1075) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{1075-1050}{\frac{218}{\sqrt{50} } }$ ) = P(Z > 0.81) = 1 - P(Z $\leq$ 0.81)

= 1 - 0.79103 = 0.20897

The above probability is calculated using z table by looking at value of x = 0.81 in the z table which have an area of 0.79103.

Therefore, probability that their mean energy consumption level for September is greater than 1075 kWh is 0.20897.

Answer from: Quest

The percent discount was 120

Answer from: Quest

{3 + i, 3 - i, 3, -1}

step-by-step explanation:

Another question on Mathematics

Mathematics, 21.06.2019 19:50

Table which ratio is equivalent to 3: 7? 6: 7 3: 21 24: 56 15: 30

Answers: 1

Answer

Mathematics, 21.06.2019 22:20

Which strategy is used by public health to reduce the incidence of food poisoning?

Answers: 2

Answer

Mathematics, 21.06.2019 23:30

Scenario: a rectangular plot of ground is 5 meters longer than it is wide. its area is 20,000 square meters. question: what equation will you find the dimensions? note: let w represent the width. options: w(w+5)=20,000 w^2=20,000+5 (w(w+5))/2=20,000 w+2(w+5)=20,000

Answers: 1

Answer

Mathematics, 22.06.2019 02:20

Which equation can be simplified to find the inverse of y = 2x2^2

Answers: 2

Answer

You know the right answer?

In one region, the September energy consumption levels for single-family homes are found to be norma...

Questions