Parity example with ASCII “2B” (hex for the “+” character): If we add an odd-parity bit as the most significant bit to the 7-bit ASCII code for 0x2B, does the resulting 8-bit pattern equal 10101011?

Introduction / Context:
Parity examples help solidify bit-level reasoning. The ASCII code 0x2B corresponds to the “+” character. Classic ASCII is a 7-bit code; many systems transmit an additional parity bit to detect errors. With odd parity, the total number of 1s in the transmitted byte (data bits + parity bit) must be odd. This question verifies that adding an odd-parity bit to ASCII 0x2B yields the pattern 10101011.

Given Data / Assumptions:

• 7-bit ASCII for “+” is 0x2B (binary 0101011 in 7 bits; 00101011 if shown as 8 bits with leading 0).
• Parity bit is placed as the most significant bit in this example.
• Odd parity means total count of 1s (including the parity bit) is odd.

Concept / Approach:
Count the number of 1s in the data bits, then choose the parity bit to make the total odd. The 7-bit 0x2B has four 1s when represented as 00101011 (the leading 0 does not change parity). Since four is even, the parity bit must be 1 to yield an odd total of five. Placing that 1 as the MSB produces 10101011.

Step-by-Step Solution:

Verification / Alternative check:
Recount ones in 10101011: there are 5 ones, which is odd. Therefore, odd parity is satisfied. If even parity had been required, the parity bit would be 0 (yielding 00101011).

Why Other Options Are Wrong:

Common Pitfalls:
Mixing 7-bit and 8-bit representations when counting ones, or forgetting which end hosts the parity bit in a specific protocol.

Final Answer:
Correct