Question 1199130
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You made an error somewhere when calculating the 1.97


The sum of the x*p(x) values is the mean.
Refer to the table below to see that the mean should be mu = 2.3 instead of 1.97
<table border = "1" cellpadding = "5"><tr><td>x</td><td>p(x)</td><td>x*p(x)</td><td>(x-mu)^2*p(x)</td></tr><tr><td>0</td><td>0.05</td><td>0</td><td>0.2645</td></tr><tr><td>1</td><td>0.2</td><td>0.2</td><td>0.338</td></tr><tr><td>2</td><td>0.15</td><td>0.3</td><td>0.0135</td></tr><tr><td>3</td><td>0.6</td><td>1.8</td><td>0.294</td></tr><tr><td>Sum</td><td>1</td><td>2.3</td><td>0.91</td></tr></table>
That value of mu is then used to calculate the column (x-mu)^2*p(x)
The sum of those values is 0.91 which is the variance. This value is exact without any rounding done to it.


Apply the square root of the variance to get the standard deviation.


SD = standard deviation
SD = sqrt(variance)
SD = sqrt(0.91)
SD = 0.953939 approximately
SD = 0.95


Various online calculators can be used to verify this result.
Here is one such calculator
<a href="https://www.mathportal.org/calculators/statistics-calculator/probability-distributions-calculator.php">https://www.mathportal.org/calculators/statistics-calculator/probability-distributions-calculator.php</a>


More practice with a similar question
<a href="https://www.algebra.com/algebra/homework/Probability-and-statistics/Probability-and-statistics.faq.question.1206570.html">https://www.algebra.com/algebra/homework/Probability-and-statistics/Probability-and-statistics.faq.question.1206570.html</a>
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