Answer:
About 34%
Explanation:
Given that :
Mean, m = 59 ; Standard deviation, s = 3
Bulb replacement numbering between 59 and 62
For, x = 59
Zscore = (x - m) / s
Z = (59 - 59) / 3 = 0
For x = 62
Z = (62 - 59) / 3 = 1
P(Z < 1) - P(Z < 0)
0.84134 - 0.5
= 0.34134
About 34%