Question

the life span of light bulbs is approximately normally distributed. The LED bulb has a mean of 2,300 days and a standard deviation of 230 days. Given a batch of 10,00 bulbs,calculate the following:

Answer 1

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:

$0.9772-0.0228=0.9544=95.44\text{\%}$

2) First let's find the values of z that corresponds to that interval:

$\begin{gathered} z_-=\frac{x_--\mu}{\sigma}=\frac{2070-2300}{230}=-1 \\ z_+=\frac{x_+-\mu}{\sigma}=\frac{2530-2300}{230}=1 \end{gathered}$

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:

$0.8413-0.1587=0.6826=68.26\text{\%}$

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:

$1-0.8413=0.1587$

And the number of bulbs would be:

$10000\cdot0.1587=1587$

Rounding to the nearest whole number, we have 1587 bulbs.

4) First let's find the z-score for 1840:

$z_{}=\frac{x_{}-\mu}{\sigma}=\frac{1840-2300}{230}=-2$

Looking at the table for z = -2, we have 0.0228, so the number of bulbs would be:

$10000\cdot0.0228=228$

So 228 bulbs would last less than 1840 days.

5) Finding the z-score for this interval, we have:

$\begin{gathered} z_-=\frac{x_--\mu}{\sigma}=\frac{2070-2300}{230}=-1 \\ z_+=\frac{x_+-\mu}{\sigma}=\frac{2760-2300}{230}=2 \end{gathered}$

The value for z = -1 is 0.1587, and the value for z = 2 is 0.9772, so we have:

$0.9772-0.1587=0.8185$

Multiplying by the number of bulbs, we have:

$10000\cdot0.8185=8185$

So the answer is 8185 bulbs.

6) The proportion was calculated above, it is 0.8185 = 81.85%