Question

1. The table shows the total annual precipitation for San Diego, California for several years. What is the mean annual precipitation to the nearest tenth of an inch? (Enter only the number.)
2. What is the standard deviation rounded to the nearest tenth?
3. In what years was the precipitation more than one standard deviation from the mean?
a) 1999 and 2002
b) 1995, 1998, and 2003
c) 1996 and 1997
d)1995, 1998, and 2002
4. The median for the data to the nearest tenth of an inch is inches.

Answer 1

@ChristopherToni

Answer 2

@dan815

Answer 3

For standard deviation, remember the nicer formula \(s = \sqrt{\dfrac{n\sum x^2 - (\sum x)^2}{n(n-1)}}\).

The x-values in this instance are the precipitation amounts.

Hence,

\(\sum x = 9.4 + 17.0 + 7.3 + 7.0 + 16.1 + 5.4 + 6.9 + 8.5 + 4.2 + 9.2 = 91\)

\(\begin{aligned}\sum x^2 &= 9.4^2 + 17.0^2 + 7.3^2 + 7.0^2 + 16.1^2 + 5.4^2 + 6.9^2 + 8.5^2 + 4.2^2 + 9.2^2 \\ &= 990.16\end{aligned}\)

Therefore, \(\displaystyle s = \sqrt{\frac{10(990.16) - 91^2}{10(10-1)}}\approx 4.24343\). Rounded to one decimal place, we see that the standard deviation is \(\boxed{s=4.2}\).

For the second part, we recall that the mean is defined by \(\overline{x} = \dfrac{\sum x}{n}\). In this case we have \(\overline{x} = \dfrac{91}{10} = 9.1\). Thus, the interval of values that are within one standard deviation of the mean is \((9.1-4.2,9.1+4.2) = (4.9,13.3) \). Thus, if \(x<4.9\) or \(x>13.3\) it is more than one standard deviation from the mean. You now want to find the values in your table that satisfy the inequalities \(x<4.9\) or \(x>13.3\). The years associated with those precipitation values will then be the answer(s) you seek.

Does this make any sense? :-)

Answer 4

yes thank you so much :)