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3 years ago

Find the median and mean of the data set below: 21,0,36,30,13,32

Mathematics

2 answers:

sineoko [7]3 years ago

6 0

im not going to give the answer but i am going to help you

Step-by-step explanation:

look the median is just the middle, if there is 2 medians then you add those 2 and then divide by 2 to get the MEDIAN in absolute.

now to find the mean all you do is simple you first add all of the numbers below then you would divide by how many numbers you have?

I hope this helps if not then have a nice day

Gnesinka [82]3 years ago

6 0

Answer:

median: 25.5

mean: 22

Step-by-step explanation:

So the median is literally the middle of the data set, but lucky for us, we have an even number, so:

21 + 30 ÷ 2 = 25.5

So the median of the set is 25.5

The mean is all the numbers added together divided by the number of terms, so:

0 + 13 + 21 + 30 + 32 + 36 = 132

132 ÷ 6 = 22

So the mean of the set is 132

hope this helps:)

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Find the value of 1<br> and 1/10-1/5

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Answer:

9/10

Step-by-step explanation:

(1 1/10) - 1/5 = (1 1/10) -2/10 = 1 + (1/10 -2/10) = 1 -1/10 = 9/10

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3 years ago

Read 2 more answers

Assume that you have a sample of n 1 equals 6, with the sample mean Upper X overbar 1 equals 50, and a sample standard deviati

tigry1 [53]

Answer:

$t=\frac{(50 -38)-(0)}{7.46\sqrt{\frac{1}{6}+\frac{1}{5}}}=2.656$

$df=6+5-2=9$

$p_v =P(t_{9}>2.656) =0.0131$

Since the p value is higher than the significance level given of 0.01 we don't have enough evidence to conclude that the true mean for group 1 is significantly higher thn the true mean for the group 2.

Step-by-step explanation:

Data given

$n_1 =6$ represent the sample size for group 1

$n_2 =5$ represent the sample size for group 2

$\bar X_1 =50$ represent the sample mean for the group 1

$\bar X_2 =38$ represent the sample mean for the group 2

$s_1=7$ represent the sample standard deviation for group 1

$s_2=8$ represent the sample standard deviation for group 2

System of hypothesis

The system of hypothesis on this case are:

Null hypothesis: $\mu_1 \leq \mu_2$

Alternative hypothesis: $\mu_1 > \mu_2$

We are assuming that the population variances for each group are the same

$\sigma^2_1 =\sigma^2_2 =\sigma^2$

The statistic for this case is given by:

$t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$

The pooled variance is:

$S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}$

We can find the pooled variance:

$S^2_p =\frac{(6-1)(7)^2 +(5 -1)(8)^2}{6 +5 -2}=55.67$

And the pooled deviation is:

$S_p=7.46$

The statistic is given by:

$t=\frac{(50 -38)-(0)}{7.46\sqrt{\frac{1}{6}+\frac{1}{5}}}=2.656$

The degrees of freedom are given by:

$df=6+5-2=9$

The p value is given by:

$p_v =P(t_{9}>2.656) =0.0131$

4 0

3 years ago

An electrical power company is looking to expand into a new market. Before they commit to supplying the new area with electricit

TiliK225 [7]

Answer:

atleast 385

Step-by-step explanation:

Given that an electrical power company is looking to expand into a new market. Before they commit to supplying the new area with electricity, they would like know the mean daily power usage for homes there.

Population std deviation = $\sigma = 50$