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Rzqust [24]
3 years ago
7

The average summer temperature in Anchorage is 69°F. If the daily temperature is normally distributed with a standard deviation

of 7°F, what percentage of the time would the temperature be between 55°F and 76°F?
Mathematics
1 answer:
mafiozo [28]3 years ago
7 0

Answer:

81.85%

Step-by-step explanation:

Given :

The average summer temperature in Anchorage is 69°F.

The daily temperature is normally distributed with a standard deviation of 7°F .

To Find:What percentage of the time would the temperature be between 55°F and 76°F?

Solution:

Mean = \mu = 69

Standard deviation = \sigma = 7

Formula : z=\frac{x-\mu}{\sigma}

Now At x = 55

z=\frac{55-69}{7}

z=-2

At x = 76

z=\frac{76-69}{7}

z=1

Now to find P(55<z<76)

P(2<z<-1)=P(z<2)-P(z>-1)

Using z table :

P(2<z<-1)=P(z<2)-P(z>-1)=0.9772-0.1587=0.8185

Now percentage of the time would the temperature be between 55°F and 76°F = 0.8185 \times 100 = 81.85\%

Hence If the daily temperature is normally distributed with a standard deviation of 7°F, 81.85% of the time would the temperature be between 55°F and 76°F.

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Answer:

A. The lowest temperature varied only 4.3 degrees from the mean.

C. No temperature varied more than 4.3 degrees from the mean.

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Step-by-step explanation:

The mean absolute deviation (MAD) is a measure of variation of the average distance between each data value and mean of a data set. It describes  the variability of values with respect to mean of a given data set.

MAD can be determined by first calculating the mean of the data, find the sum of absolute value of the difference between each datum and the mean. Then divide this sum by the number of datum of the given data.

With Hui;s record and MAD for his data set, he would be able to describe the varations between the mean and each datum.

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Suppose that 600 yards of fencing material are available to fence in
postnew [5]

Answer:

Length = 150 yards

Width = 100 yards

Step-by-step explanation:

We want 600 yards of fencing that will result in the largest 2 fenced corrals, sharing a common border.

It will take the shape of a rectangle, with a dividing fence down the center.

Let W and L,  Width and Length of the larger enclosure.

See attachment.

W= Area of the larger enclosure.

The perimeter is 2W + 2L.

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---

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---

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Area = W x L

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