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Lina20 [59]
2 years ago
8

The average age of a breed of dog is 19.5 years. if the distribution of their ages is normal and 20% of dogs are older than 22.6

years, find the standard deviation.
SAT
1 answer:
viva [34]2 years ago
3 0

The  standard deviation of the average age of a breed of dog  is said to be 3.69.

<h3>What is the standard deviation about?</h3>

Note that we take:

  • x = dog age has mean 19.5
  • Unknown standard deviation.

Hence:

z = (x - mean) /SD is distributed N (0,1) - mean o standard deviation 1 (standard normal)

Since 20% of dogs are known to be older than 22.6 years:

So 22.6 years is said to be 80th percentile,

Since z matching to 80th percentile (using the standard normal table) s is <em>0.84.</em>

Then one need to plug z, x which is  = 22.6

mean = 19.5 into the stated z equation:

Hence,

0.84 = (22.6 - 19.5)/SD

SD = (22.6 - 19.5)/0.84

= 3.69

Learn more about standard deviation from

brainly.com/question/475676

#SPJ1

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The pH shows the degree of acidity of alkalinity of a solution. The pH of the HF can be obtained from the ICE table.

<h3>What is pH?</h3>

The term pH refers to the hydrogen ion concentration of a solution. Recall that acids produce hydrogen or oxonium ion in solution.

The first step to determine the pH of the HF solution is to set up the ICE table as shown;

             HF + H2O  ⇄ H3O^+ + F^-

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The values of the initial concentration of HF  as well as the Ka of HF are not given. If these two values are give, the pH can be obtained from;

Ka = [H3O^+] [ F^-]/[HF] Knowing that  [H3O^+] = [ F^-] = x

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Find the residual values, and use the graphing calculator tool to make a residual plot.
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Answer:

  • No, the points are evenly distributed about the x-axis.

Explanation:

<u>1. Write the table with the data:</u>

  x      given     predicted       residual

  1       - 3.5         - 1.1

  2      - 2.9           2

  3       - 1.1            5.1

  4        2.2           8.2

  5        3.4           1.3

<u>2. Complete the column of residuals</u>

The residual is the observed (given) value - the predicted value.

  • residual = given - predicted.

Thus, the complete table, with the residual values are:

  x      given     predicted       residual

  1       - 3.5         - 1.1                 - 2.4

  2      - 2.9           2                  - 4.9

  3       - 1.1            5.1                - 6.2

  4        2.2           8.2               - 6.0

  5        3.4           1.3                   2.1

<u>3. Residual plot</u>

You must plot the last column:

  x      residual

  1       - 2.4

  2      - 4.9

  3       - 6.2

  4       - 6.0

  5        2.1

See the plot attached.

<em>Does the residual plot show that the line of best fit is appropriate for the data?</em>

Ideally, a residual plot for a line of best fit that is appropiate for the data must not show any pattern; the points should be randomly distributed about the x-axis.

But the points of the plot are not randomly distributed about the x-axis:  there are 4 points below the x-axis and 1 point over the x-axis: there are more negative residuals than positive residuals.  This is a pattern. Also, you could say that they show a curve pattern, which drives to the same conclusion: the residual plot shows that the line of best fit is not appropiate for the data.

Thus, the conclusion should be: No, the points have a pattern.

  • 1. "<em>Yes, the points have no pattern</em>": false, because as shown, the points do have a pattern, which makes the residual plots does not show that the line of best fit is appropiate for the data.

  • 2. "<em>No, the points are evenly distributed about the x-axis</em>": true. As already said the points have a pattern. It is a curved pattern, and this <em>shows the line of best fit is not appropiate for the data.</em>

  • 3. "<em>No, the points are in a linear pattern</em>": false. The points are not in a linear pattern.

  • 4. "<em>Yes, the points are in a curved pattern</em>": false. Because the points are in a curved pattern, the residual plot shows that the line of best fit is not appropiate for the data.

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(Copied from google)

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