P(1,100<X<1,300)=P(z1<z<z2),
where z1 and z2 are z-scores of 1,100 and 1,300.
Let's find them.
z=σx−μ; μ=1,214; σ=122.
z1=1221,100−1,214≈−0.93,
z2=1221,300−1,214≈0.70.
So we have to rewrite the probability ang use z-table:
P(1,100<X<1,300)=P(−0.93<z<0.70)=
=0.7580−0.1762=0.5818=58.18%.
Answer: 58.18%.
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