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Normalizing the mantissa in floating point representation

I went through a question which asked to represent $0.145 * 2^{14}$ in normalized floating point arithmetic with the format

1 - Sign bit
7 - Exponent in Excess-64 form
8 - Mantissa

$(0.145)_{10} = (0.00100101001...)_2$ (say A). We shift it 3 bits to left to make it normalized $(1.00101001)_2 * 2^{11}$.

Exponent = $11+64 = (75)_{10} = (1001011)_2$ and Mantissa = $(01001001)_2$.

So floating point representation should be $(0\;1001011\;00101001)_2 = (4B29)_{16}$.

But the solution considered $(0.145)_{10} = (0.00100101)_2$(upto 8 bits only). Shifting it to left by inserting zeroes to the right makes it $(01001000)_2$ instead of $(01001001)_2$. And the representation becomes $(4B28)_{16}$.

So while normalizing, does the processor takes into account the mantissa bits beyond 8 bits too? Or just rounds it off?

Does it store the mantissa in fixed point representation? How does it all work?