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_-=(x_--\mu)/(\sigma)=(2070-2300)/(230)=-1 \\ z_+=(x_+-\mu)/(\sigma)=(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}=(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_-=(x_--\mu)/(\sigma)=(2070-2300)/(230)=-1 \\ z_+=(x_+-\mu)/(\sigma)=(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%