The standard deviation of the length of this plant be 25.
m = 60(mean)
Let the standard deviation be s,
As, the probability for 50<x<70 be 0.75
Z = (x−m)/s
for x=50,
Z=0
for x=70,
Z=(70−50)/s=20/s
Now, the probability for 50<x<70 be,
P(50<x<70) = 0.75
In the standard form 0<z<(20/s)
using Z-table, the area from 0 to 20/s,
= 0.75
s = 25
Hence, the standard deviation of the length of this plant be 25.
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