SOLUTION: Given the following table: y | 15 30 60 90 p(y)| .2 .3 .4 .1 Find: Mean, Standard deviation, and probability that the time is within 1 SD of its mean. Any help woul

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Question 1161365: Given the following table:
y | 15 30 60 90
p(y)| .2 .3 .4 .1
Find: Mean, Standard deviation, and probability that the time is within 1 SD of its mean.
Any help would be appreciated, thanks!
Answer by jim_thompson5910(35256)   (Show Source): You can put this solution on YOUR website!

Answers:
Mean = 45
Standard deviation = 23.2379000772446 (approximate; round however you need to)
Probability the time is within one standard deviation of the mean = 0.7

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Work Shown:

Part (a): Finding the mean

To find the mean, we multiply each y value by its corresponding probability p(y) value as shown in the table below
yp(y)y*p(y)
150.23
300.39
600.424
900.19

Example: in row 1 we have y*p(y) = 15*0.2 = 3

Add up the values in the y*p(y) column to get the mean
mean = 3+9+24+9 = 45

We will use the mean later when it comes to calculating the standard deviation

------------------------------------------

Part (b): Finding the standard deviation

First we must construct a new column labeled y^2*p(y). This is a column of values where we square the y values, and then multiply those squares by their corresponding p(y) values. For instance, row 1 has y = 15, so y^2 = 15^2 = 225, which leads to y^2*p(y) = 225*0.2 = 45. It's a coicidence that this also happens to be the value of the mean. The other rows are computed in a similar fashion.

Here's the updated table
yp(y)y*p(y)y^2*p(y)
150.2345
300.39270
600.4241440
900.19810

We then add up everything in that new y^2*p(y) column which yields
45+270+1440+810 = 2565

Next, we square the mean.
mu = mean
mu = 45 found earlier
mu^2 = 45^2
mu^2 = 2025

Subtract this result off the sum of the y^2*p(y) column to get the variance sigma^2
sigma^2 = (sum of y*p(y) column) - ( mu^2 )
sigma^2 = 2565 - 2025
sigma^2 = 540

Lastly, we apply the square root to both sides to get the value of sigma, which is the standard deviation
sigma^2 = 540
sigma = sqrt(540)
sigma = 23.2379000772446 which is approximate

We will use this for part (c), along with the mean as well.

------------------------------------------

Part (c): probability that the time is within 1 SD of its mean

From part (b), we found the approximate standard deviation to be 23.2379000772446

Add and subtract this from the mean to get the values of U and L respectively
L = lower value that is 1 std dev from the mean
L = mu-sigma
L = 45 - 23.2379000772446
L = 21.7620999227553 which is approximate

U = upper value that is 1 std dev from the mean
U = mu+sigma
U = 45 + 23.2379000772446
U = 68.2379000772447

The values of L and U were approximately 21.7620999227553 and 68.2379000772447

Looking at the table, we are searching for y values that are between those L and U values.
This applies for y = 30 and y = 60 only
In other words, y = 30 and y = 60 make the inequality 21.7620999227553 < y < 68.2379000772447 true.

So from here, we add up the corresponding probabilities for those y values
0.3 + 0.4 = 0.7

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