(A)
Here we have given that
= population mean =150
= population standard deviation=5
Now we want to find
The Probablity that the machine will dispense between 150 and 153 mg
i.e P(150 < X < 154)
For that 1st we want to find the Zscore
For X=150
=
=0.00
Now, For x= 154
Zscore =
=
=-0.80
i.e we get
=0.2119 – 0.5000 ( using z standard normal talble)
=0.2881
That is
We get
P(150 < X < 154) =0.2881
that is here option A is correct
(B)
Here we have given that
= population mean =150
= population standard deviation=5
Now we want to find
The Probablity that the temperature of the coffee macchiato is more than 164 Fahrenheit.
i.e P( X> 164)
For that 1st we want to find the Zscore
For X=164
=
=2.80
i.e we get
P( Z > 2.80) =1-P( Z < 2.80)
=1-0.9974 ( using standard normal Z table)
=0.0026
i.e we get
P(Z > 164) = 0.0026
That is here option C is correct.
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