Respuesta :
Answer:
The probability that the mean battery life would be greater than 533.2 minutes (in a sample of 75 batteries) is [tex] \\ P(z>0.48) = P(x>533.2) = 0.3156[/tex]
Step-by-step explanation:
The main thing we have to take into account in this question is that we are about to find the probability of a sample mean. The distribution for sample means follows a normal distribution with mean [tex] \\ \mu[/tex] and standard deviation [tex] \\ \frac{\sigma}{\sqrt{n}}[/tex]. Mathematically
[tex] \\ \overline{x} \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]
For values of the sample [tex] \\ n \ge 30[/tex], no matter the distribution the data come from.
And the variable z follows a standard normal distribution, and, as we can remember, this distribution has a mean = 0 and a standard deviation = 1. Mathematically
[tex] \\ z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex] [1]
That is
[tex] \\ z \sim N(0, 1)[/tex]
We have a variance of 3364. That is, a standard deviation of
[tex] \\ \sigma^2 = 3364; \sigma = \sqrt{3364} = 58[/tex]
The population mean is
[tex] \\ \mu = 530[/tex]
The sample size is [tex] \\ n = 75[/tex]
The sample mean is [tex] \\ \overline{x} = 533.2[/tex]
With all this information, we can solve the question
The probability that the mean battery life would be greater than 533.2 minutes
Using equation [1]
[tex] \\ z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
[tex] \\ z = \frac{533.2 - 530}{\frac{58}{\sqrt{75}}}[/tex]
[tex] \\ z = \frac{3.2}{\frac{58}{8.66025}}[/tex]
[tex] \\ z = \frac{3.2}{6.69726}[/tex]
[tex] \\ z = 0.47780[/tex]
With this value of z we can consult a cumulative standard normal table (or use some statistic program) to find the cumulative probability for z (and remember that this variable follows a standard normal distribution).
Most standard normal tables have values for z for only two decimals, so we can round the previous value for z as z = 0.48.
Then
[tex] \\ P(z<0.48) = 0.68439[/tex]
However, in the question we are asked for [tex] \\ P(z>0.48) = P(x>533.2)[/tex]. As well as all normal distributions, the standard normal distribution is symmetrical around the mean, and we have
[tex] \\ P(z>0.48) = 1 - P(z<0.48)[/tex]
Thus
[tex] \\ P(z>0.48) = 1 - 0.68439[/tex]
[tex] \\ P(z>0.48) = 0.31561[/tex]
Rounding to four decimal places, we have
[tex] \\ P(z>0.48) = 0.3156[/tex]
So, the probability that the mean battery life would be greater than 533.2 minutes is (in a sample of 75 batteries) [tex] \\ P(z>0.48) = P(x>533.2) = 0.3156[/tex].
