(a) the time for the event of the highest 5% is 15.29 minutes.
(b) the time for the event of the lowest 50% is 12 minutes.
(c) the time for the event of the middle 95% is 8.08 minutes and 15.92 minutes.
(d) the time for the event of the lowest 80% is 13.68 minutes.
Step-by-step explanation:
We are given that the time required to cook a pizza at a neighborhood pizza joint is normally distributed with a mean of 12 minutes and a standard deviation of 2 minutes.
Let X = the time required to cook a pizza at a neighborhood pizza joint.
The z-score probability distribution for the normal distribution is given by;
               Z  =   ~ N(0,1)
where, = mean time = 12 minutes
      = standard deviation = 2 minutes
(a) We have to find the time for the event of the highest 5%, that means;
      P(X > x) = 0.05   {where x is the required time}
      P( > ) = 0.05
      P(Z > ) = 0.05
Now, in the z table, the critical value of x that represents the top 5% of the area is given as 1.645, i.e;
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           x = 12 + 3.29 = 15.29 minutes.
Hence, the time for the event of the highest 5% is 15.29 minutes.
(b) We have to find the time for the event of the lowest 50%, that means;
      P(X < x) = 0.50   {where x is the required time}
      P( < ) = 0.50
      P(Z < ) = 0.50
Now, in the z table, the critical value of x that represents the lowest 50% of the area is given as 0, i.e;
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           x = 12 + 0 = 12 minutes.
Hence, the time for the event of the lowest 50% is 12 minutes.
(c) We have to find the time for the event of the middle 95%, that means we have to find the time for the event of 2.5% and above 97.5%;
Firstly, the time for the event of the lowest 2.5%, i.e;
      P(X < x) = 0.025   {where x is the required time}
      P( < ) = 0.025
      P(Z < ) = 0.025
Now, in the z table, the critical value of x that represents the lowest 2.5% of the area is given as -1.96, i.e;
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           x = 12 - 3.92 = 8.08 minutes.
Firstly, the time for the event of the below 97.5%, i.e;
      P(X < x) = 0.975   {where x is the required time}
      P( < ) = 0.975
      P(Z < ) = 0.975
Now, in the z table, the critical value of x that represents the top 2.5% of the area is given as 1.96, i.e;
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           x = 12 + 3.92 = 15.92 minutes.
Hence, the time for the event of the middle 95% is 8.08 minutes and 15.92 minutes.
(d) We have to find the time for the event of the lowest 80%, that means;
      P(X < x) = 0.80   {where x is the required time}
      P( < ) = 0.80
      P(Z < ) = 0.80
Now, in the z table, the critical value of x that represents the lowest 80% of the area is given as 0.8416, i.e;
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           x = 12 + 1.68 = 13.68 minutes.
Hence, the time for the event of the lowest 80% is 13.68 minutes.