Question

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. Find the time for each event.

Answer 1

Answer:

(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  =  $\frac{X-\mu}{\sigma}$  ~ N(0,1)

where, $\mu$ = mean time = 12 minutes

           $\sigma$ = 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( $\frac{X-\mu}{\sigma}$ > $\frac{x-12}{2}$ ) = 0.05

           P(Z > $\frac{x-12}{2}$ ) = 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;

                      $\frac{x-12}{2}=1.645$

                      ${x-12}=1.645\times 2$

                      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( $\frac{X-\mu}{\sigma}$ < $\frac{x-12}{2}$ ) = 0.50

           P(Z < $\frac{x-12}{2}$ ) = 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;

                      $\frac{x-12}{2}=0$

                      ${x-12}=0$

                      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( $\frac{X-\mu}{\sigma}$ < $\frac{x-12}{2}$ ) = 0.025

           P(Z < $\frac{x-12}{2}$ ) = 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;

                      $\frac{x-12}{2}=-1.96$

                      ${x-12}=-1.96 \times 2$

                      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( $\frac{X-\mu}{\sigma}$ < $\frac{x-12}{2}$ ) = 0.975

           P(Z < $\frac{x-12}{2}$ ) = 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;

                      $\frac{x-12}{2}=1.96$

                      ${x-12}=1.96 \times 2$

                      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( $\frac{X-\mu}{\sigma}$ < $\frac{x-12}{2}$ ) = 0.80

           P(Z < $\frac{x-12}{2}$ ) = 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;

                      $\frac{x-12}{2}=0.8416$

                      ${x-12}=0.8416 \times 2$

                      x = 12 + 1.68 = 13.68 minutes.

Hence, the time for the event of the lowest 80% is 13.68 minutes.