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Sati [7]
3 years ago
14

The per capita electric power consumption level in a recent year in Ecuador is normally distributed, with a mean of 471.5 kilo-w

att hours and a standard deviation of 187.9 kilowatt-hours. Random samples of size 35 are drawn from this population. Find (a) the mean and (b) the standard deviation of the sampling distribution of sample means. Round the answer from part (b) to the third decimal place.
Mathematics
1 answer:
Olin [163]3 years ago
7 0

Answer:

a) 471.5 kilo-watt hours.

b) 31.76 kilo-watt hours

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For the population:

Mean 471.5 kilo-watt hours.

Standard deviation of 187.9 kilowatt-hours.

For the sample:

Sample size of 35, by the Central Limit Theorem:

a) Mean

471.5 kilo-watt hours.

b) Standard deviation

s = \frac{187.9}{\sqrt{35}} = 31.76

31.76 kilo-watt hours

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A newspaper poll asked respondents if they trusted "eco friendly" labels on cleaning products. Out of 1000 adults surveyed, 498
Paul [167]

Answer:

So the p value obtained was a very high value and using the significance level assumed \alpha=0.05 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of respondents that trust these labels is not significanlty less than 0.5 or 50%.  

Step-by-step explanation:

1) Data given and notation

n=1000 represent the random sample taken

X=498 represent the adults that trust these labels

\hat p=\frac{498}{1000}=0.498 estimated proportion of respondents that trust these labels

p_o=0.5 is the value that we want to test

\alpha represent the significance level

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the proportion of respondents that trust these labels is at least 50%:  

Null hypothesis:p\geq 0.5  

Alternative hypothesis:p < 0.5  

When we conduct a proportion test we need to use the z statisitc, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

3) Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.498 -0.5}{\sqrt{\frac{0.5(1-0.5)}{1000}}}=-0.126  

4) Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level is not provided but we can assume it as \alpha=0.05. The next step would be calculate the p value for this test.  

Since is a left tailed test the p value would be:  

p_v =P(Z  

So the p value obtained was a very high value and using the significance level assumed \alpha=0.05 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of respondents that trust these labels is not significanlty less than 0.5 or 50%.  

7 0
3 years ago
Super easy plz answer all for 20 points just adding
sergejj [24]
7/8+9/10=71/40=1 31/40
1/4+2/3=11/12
1/2+1/4 is different from 1/2+1/3 because 
1/2+1/4= 2/4+1/4=  5/1 =1  1/4
1/2+1/3=3/6+2/6=5/6
6 0
3 years ago
The dimensions of the tank are 50 cm by 32 cm by 20 cm. The tank is full of water and sand. The ratio of the volume of water to
Sauron [17]

Given:

Dimensions of a tank are 50 cm by 32 cm by 20 cm

The tank is full of water and sand

Ratio of volume of water to the volume of sand is 5:1

To find:

Volume of water in the tank in liters

Solution:

Without loss of generality, let,

Length of tank = 50 cm

Breadth of tank = 32 cm

Height of tank = 20 cm

Then, we have,

Total Volume of the tank = Length * Breadth * Height

\Rightarrow Total Volume = 50*32*20 cc

\Rightarrow Total Volume = 32000 cc

It is given that,

Volume of water : Volume of sand = 5:1

\Rightarrow (Volume of water)/(Volume of sand) = 5/1

Cross multiplying, we have,

Volume of sand = (Volume of water)/5

It is also given that the tank is full of water and sand. This implies that,

Volume of water + Volume of sand = Total Volume

Using the value, we obtained, we get,

Volume of water + (Volume of water)/5 = Total Volume

\Rightarrow (6 * Volume of water)/5 = Total Volume

Plugging in the value of total volume, we get,

\Rightarrow (6 * Volume of water)/5 = 32000 cc

\Rightarrow Volume of water = \frac{32000*5}{6} cc

\Rightarrow Volume of water \approx 26,666.6667 cc

We know that,

1 cc = 0.001 liters

\Rightarrow 26,666.6667 cc = (26666.6667 * 0.001) liters

\Rightarrow 26,666.6667 cc \approx 26.667 liters

Thus, Volume of water in the tank is approximately 26.667 liters

Final Answer:

Volume of water in the tank (in liters) is approximately 26.667 liters

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2 years ago
3/5 of the 30 students are boys how many students are girls
anygoal [31]

if 3/5 are boys

1- 3/5

5/5 -3/5 = 2/5  are girls

2/5 * 30 = 60/5 = 12

12 are girls

6 0
2 years ago
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How many centimeters are in 5.2 inches? [1 inch = 2.5 cm]
Molodets [167]
This would be about 13 cm
6 0
3 years ago
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