Solution:-

Let us consider the assumed mean A = \[64\]. Here h = \[1\]

Mean,

\[\bar{X}=A+\frac{\sum\limits_{i=1}^{a}{{{f}_{i}}{{y}_{i}}}}{N}\times h\]

Where A = \[64\], h = \[1\]

\[\overline{x}=64+((0/100)\times 1)\]

= \[64+0\]

= \[64\]

Then, variance,

\[{{\sigma }^{2}}=\frac{{{h}^{2}}}{{{N}^{2}}}[N\sum{{{f}_{i}}y_{i}^{2}-(\sum{{{f}_{i}}{{y}_{i}}{{)}^{2}}]}}\]

\[{{\sigma }^{2}}=({{1}^{2}}/{{100}^{2}})[100(286)-{{0}^{2}}]\]

= \[(1/10000)[28600-0]\]

= \[28600/10000\]

= \[2.86\]

Hence, standard deviation = \[\sigma =\sqrt{2.886}\]

= \[1.691\]

Therefore, Mean = \[64\] and Standard Deviation = \[1.691\]