Question

A safety light is designed so that the times between flashes are normally distributed with a mean of 2.50s and a standard deviation of 0.50s. a. Find the probability that an individual time is greater than 3.00 s. b. Find the probability that the mean for 70 randomly selected times is greater than 3.00 s. c. Given that the light is intended to help people see an​ obstruction, which result is more relevant for assessing the safety of the​ light?

Answer 1

Answer:

a) $P(X>3.00)=P(Z>1)=1-P(Z$

b) $P(\bar X >3.00)=P(Z>8.37)=1-P(Z$

c) Part (b) is more significant because the average time for a sample of size 70 is important

Step-by-step explanation:

1) Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

2) Part a

Let X the random variable that represent the times between flashes of a population, and for this case we know the distribution for X is given by:

$X \sim N(2.5,0.5)$  

Where $\mu=2.5$ and $\sigma=0.5$

We are interested on this probability

$P(X>3.00)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X>3.00)=P(\frac{X-\mu}{\sigma}>\frac{3.00-\mu}{\sigma})=P(Z>\frac{3.00-2.5}{0.5})=P(Z>1)$

And we can find this probability on this way:

$P(Z>1)=1-P(Z$

3) Part b

From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And we want this probability

$P(\bar X >3.00)=P(Z>\frac{3.00-2.5}{\frac{0.5}{\sqrt{70}}}=8.37)$

And using a calculator, excel ir the normal standard table we have that:

$P(Z>8.37)=1-P(Z$

4) Part c

Part (b) is more significant because the average time for a sample of size 70 is important