1) In order to find the proportion of bulbs that are expected to last within 2 standard deviations of the mean, we need to find the value in the z-table that corresponds to z = -2 and z = 2, then find the difference of these values.
For z = -2, we have 0.0228, and for z = 2 we have 0.9772, so the difference is:
![0.9772-0.0228=0.9544=95.44\text{\%}](https://img.qammunity.org/2023/formulas/mathematics/college/spchhkb7jp3dneb2p3mk8sw8hqr8ceve01.png)
2) First let's find the values of z that corresponds to that interval:
![\begin{gathered} z_-=(x_--\mu)/(\sigma)=(2070-2300)/(230)=-1 \\ z_+=(x_+-\mu)/(\sigma)=(2530-2300)/(230)=1 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/wy1t3wikr6amdcjz42zhyg1s1zp9ywh9mz.png)
Looking at the z-table for z = -1 and z = 1, we have 0.1587 and 0.8413, so the probability of the interval is:
![0.8413-0.1587=0.6826=68.26\text{\%}](https://img.qammunity.org/2023/formulas/mathematics/college/22f3f0lwbqgaalj691wfcvc2fx2fboykq7.png)
3) To find how many will last longer than 2530, we know that this value corresponds to z = 1, so the value in the table is 0.8413.
The percentage of bulbs that will last longer than this is:
![1-0.8413=0.1587](https://img.qammunity.org/2023/formulas/mathematics/college/x3trjx0svelhtdah1tj1qv1gsoqok75y4b.png)
And the number of bulbs would be:
![10000\cdot0.1587=1587](https://img.qammunity.org/2023/formulas/mathematics/college/xn8g554vbatlfqdehptxf26vxn4n0u3hqz.png)
Rounding to the nearest whole number, we have 1587 bulbs.
4) First let's find the z-score for 1840:
![z_{}=\frac{x_{}-\mu}{\sigma}=(1840-2300)/(230)=-2](https://img.qammunity.org/2023/formulas/mathematics/college/nndd4jhki5xcw4o3ri57hxayd643z0tbpg.png)
Looking at the table for z = -2, we have 0.0228, so the number of bulbs would be:
![10000\cdot0.0228=228](https://img.qammunity.org/2023/formulas/mathematics/college/uc2wdzylypbzeodtxlcegy8djnpljqmsz9.png)
So 228 bulbs would last less than 1840 days.
5) Finding the z-score for this interval, we have:
![\begin{gathered} z_-=(x_--\mu)/(\sigma)=(2070-2300)/(230)=-1 \\ z_+=(x_+-\mu)/(\sigma)=(2760-2300)/(230)=2 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/h2e9b2py80kkzx8brg85i9hczhytagiecq.png)
The value for z = -1 is 0.1587, and the value for z = 2 is 0.9772, so we have:
![0.9772-0.1587=0.8185](https://img.qammunity.org/2023/formulas/mathematics/college/9xry91ybtfhuqn1wktc7spe4sq65qjsdtr.png)
Multiplying by the number of bulbs, we have:
![10000\cdot0.8185=8185](https://img.qammunity.org/2023/formulas/mathematics/college/l7njsmibdep4o0qiqz81bsj3abckc15g9v.png)
So the answer is 8185 bulbs.
6) The proportion was calculated above, it is 0.8185 = 81.85%