If X is the number of marks required to be included in the 80%, and the mean is 55 with SD=10, we have the z score=(X-55)/10. Any score below X puts the student in the excluded 20%. The value of z corresponding to 0.2 is approximately -0.84, so X-55=-8.4, making X=46.6 or 47 to the nearest whole mark.

Put X=46.6; z score=(46.6-55)/10=-0.84. P(-0.84)=1-P(0.84)=1-0.2=0.8=80% according to normal distribution tables.

P(-0.84) is the probability of the marks being below the "pass" mark of 47 so that 80% of students will have sufficient marks for promotion. I looked up the z score for 80% or 0.8 and discovered that it corresponded roughly to z=0.84, so the z score below which the marks would not qualify for promotion is z=-0.84, the lower 20%. That's where I got -0.84 from.