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Alenkasestr [34]
4 years ago
8

246, 299, 360, 404, 379, 199, 279, 749, 794, 849, 914

Mathematics
1 answer:
Sholpan [36]4 years ago
5 0

Answer:

Mean = 497.5

Median = 379.0

First Quartile = 289

Third Quartile = 771.5

Step-by-step explanation:

Mean is used to measure the central tendency of data which represents the whole data in the best way. It can be found as the ratio of the sum of all the observations to the total number of observations.  

⇒ Mean=\frac{246+ 299+ 360+ 404+ 379+ 199+ 279+ 749+ 794+ 849+ 914}{11}

⇒ Mean = 497.5

Median is the middle observation of given data. It can be found by following steps:

Arranging data in ascending or descending order.

Taking the average of middle two value if the total number of observation is even, and this average is our median.

or, if we odd number of observation then the most middle value is our median.  

Here, number of observation is 11.

So the middle value is (11+1)÷2 = 6th term

⇒ Median = 379

The mode is the observation which has a high number of repetitions (frequency).

Here frequency of all observation is same. So, it is multi- modal data.

First Quartile is the middle value between Minimum value and Median of data after arranging data in ascending order.

First Quartile (Q₁) = 289

The third Quartile is the middle value between Median and Maximum Value of data after arranging data in ascending order.

Third Quartile (Q₃) = 771.5

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Answer:

a) \bf 0.3446^{25}=2.7095*10^{-12}

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Step-by-step explanation:

a)

Since 4300/25 = 172, in average every single person should weight 172 pounds to exceed the design limit.

The probability that one person weights 172 pounds or more is the area under the Normal curve with mean 160 pounds and standard deviation 30 pounds to the right of 172.

<em>In Excel and OpenOffice Calc, this value is found with the formula </em>

<em>=1-NORMDIST(172;160;30;1) </em>

<em> (NORMDIST(172;160;30;1) gives the area to the left of 172, so 1-NORMDIST(172;160;30;1) gives the area to the right of 172) </em>

and equals 0.3446

(see picture 1)

Hence, the probability that all the 25 people exceeds 172 pounds equals

\bf 0.3446^{25}=2.7095*10^{-12}

b)

Similarly, we must find a weight w such that if p is the area to the left of w, then  

\bf p^{25}>0.0001  

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\bf p=\sqrt[25]{0.0001}=0.6918

w would be the point such that the area under the Normal curve with mean 160 pounds and standard deviation 30 pounds to the right of w equals 0.6918.

<em>In Excel and OpenOffice Calc this is found with </em>

<em>=NORMINV(1-0.6918;160;30) </em>

and equals 144.97 pounds

(See picture 2)

and the design limit that is exceeded by 25 occupants with probability 0.0001 is 25*144.97 = 3,624.25 pounds

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4 years ago
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