Results_precip_SLP_TeddyAllen

=Precipitation and Sea Level Pressure (SLP)= Te(d^2)y A(l^2)en

=2. precip and SLP means and variances: space, time, spacetime= The **GRAND (space+time) mean** of precipitation and SLP are : **4.74mm; 1010.97hPa.** mean(x)

The **GRAND (space+time) variance** of precipitation and SLP are: ** 12.63 (mm)^2; 2.37(hPa)^2 ** variance = (sum(x^2) - sum(x)^2) /N

The **GRAND standard deviations** are: **3.55mm; 1.54hPa** sqrt(var_n(x))

The **SPATIAL variances** of my TIME MEAN longitude section are: **5.76 (mm)^2; 1.16 (hPa)^2** var_n( total(x,2)/NMONS )

The **TEMPORAL variances** of my LONGITUDE MEAN time series are: **0.94 (mm)^2; 0.55 (hPa)^2** var_n( total(x,1)/NLONS ) =** 3. the assignment section **= 1. Confirm that the time mean of the anomalies as defined above is 0 2. Is the spatial mean of the anomalies (as defined above) 0? **The RED time-series plots below show that the spatial mean of the anomalies (longitude mean) is NOT zero.** Is it the same as the time series of the spatial mean of the raw data? Or is it a new thing? **The spatial mean of the anomalies is distinct from the actual spatial mean of the raw data .**
 * The above statement is confirmed by averaging the anomalies for each longitude grid. **

__ TO COMPUTE THE BELOW TO PLOT: __ spatial (longitude) mean of the entire data series lonmean_x = (total(x,1)/nlons) spatial (longitude) mean of the anomaly lonmean_anomx = (total(anomx,1)/nlons) difference between the series and anomaly diffx = (lonmean_x - lonmean_anomx)