11.
We shall obtain the values of z from the standard normal tables.
a)
z=1.92
b)
z=2.22
c)
z=-0.5
d)
If the area between -z and z is 0.9282 it implies that the area to the left of -z is 21−0.9282=0.0359. Thus −z=−1.8⟹z=1.8
12.
μ=82σ=4.8
a)
p(x<89.2)=p(Z<4.889.2−82)=p(Z<1.5)=ϕ(1.5)=0.9332
b)
p(x>78.4)=p(Z>4.878.4−82)=p(Z>−0.75)=1−p(Z<−0.75)=1−0.2266=0.7734
c)
p(83.2<x<88)=p(4.883.2−82<Z<4.888−82)=p(0.25<Z<1.25)=ϕ(1.25)−ϕ(0.25)=0.8944−0.5987=0.2957
d)
p(73.6<x<90.4)=p(4.873.6−82<Z<4.890.4−82)=p(−1.75<Z<1.75)=ϕ(1.75)−ϕ(−1.75)=0.9599−0.0401=0.9198
13.
μ=55.8σ=12.2
a)
p(x<49.7)=p(Z<12.249.7−55.8)=p(Z<−0.5)=0.3085
b)
p(61.9<x<74.1)=p(12.261.9−55.8<Z<12.274.1−55.8)=p(0.5<Z<1.5)=ϕ(1.5)−ϕ(0.5)=0.9332−0.6915=0.2417
c)
p(x>86.3)=p(Z>12.286.3−55.8)=p(Z>2.5)=1−p(Z<2.5)=1−0.9938=0.0062
14.
Here, we find the value y such that p(x<y)=0.90
Now,
p(x<y)=p(Z<12.2y−55.8)=0.90. This implies that, ϕ(12.2y−55.8)=0.90⟹12.2y−55.8=Z0.90 where, Z0.90 is the table value associated with 0.90. From tables, Z0.90=1.281552
So,
12.2y−55.8=1.281552⟹y=71.4349344
Therefore, the probability that one can assemble the desk in less than 71.4349344 minutes is 0.90
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