The number of arrivals per minute at a bank located in the central business district of a large city was recorded over a period of 200 minutes, with the results shown in the table below. Complete (a) through (c) to the right.
\begin{tabular}{cc}
Arrivals & Frequency \\
\hline 0 & 21 \\
1 & 35 \\
2 & 4 \\
3 & 36 \\
4 & 28 \\
5 & 22 \\
6 & 11 \\
7 & 5 \\
8 & 2
\end{tabular}
a. Compute the expected number of arrivals per minute.
$\mu=2.81$
(Type an integer or decimal rounded to three decimal places as needed.)
b. Compute the standard deviation.
\[
\sigma=\square
\]
(Type an integer or decimal rounded to three decimal places as needed.)

Step 1 ：First, we need to compute the expected number of arrivals per minute. This is done by multiplying each arrival number by its frequency, summing up these products, and then dividing by the total number of minutes. The total number of minutes is 200.

Step 2 ：The sum of the products of arrivals and their frequencies is: \(0*21 + 1*35 + 2*4 + 3*36 + 4*28 + 5*22 + 6*11 + 7*5 + 8*2 = 0 + 35 + 8 + 108 + 112 + 110 + 66 + 35 + 16 = 490\)

Step 3 ：So, the expected number of arrivals per minute is \(\frac{490}{200} = 2.45\)

Step 4 ：Next, we need to compute the standard deviation. To do this, we first need to compute the variance. The variance is the average of the squared differences from the Mean.

Step 5 ：The squared differences from the Mean are: \((0-2.45)^2*21 + (1-2.45)^2*35 + (2-2.45)^2*4 + (3-2.45)^2*36 + (4-2.45)^2*28 + (5-2.45)^2*22 + (6-2.45)^2*11 + (7-2.45)^2*5 + (8-2.45)^2*2\)

Step 6 ：This simplifies to: \(5.0025*21 + 2.0025*35 + 0.2025*4 + 0.3025*36 + 2.4025*28 + 6.5025*22 + 12.6025*11 + 20.7025*5 + 30.8025*2 = 105.0525 + 70.0875 + 0.81 + 10.89 + 67.27 + 143.055 + 138.6275 + 103.5125 + 61.605 = 700.91\)

Step 7 ：So, the variance is \(\frac{700.91}{200} = 3.50455\)

Step 8 ：Finally, the standard deviation is the square root of the variance, which is \(\sqrt{3.50455} = 1.872\)

Step 9 ：So, the expected number of arrivals per minute is \(\boxed{2.45}\) and the standard deviation is \(\boxed{1.872}\)