Respuesta :
Answer:
a) [tex]\bar X \sim N(\mu=30, \frac{2}{\sqrt{16}})[/tex]
b) [tex]Se=\frac{\sigma}{\sqrt{n}}=\frac{2}{\sqrt{16}}=0.5[/tex]
c) [tex]P(\bar X >31.25)=0.006=0.6\%[/tex]
d) [tex]P(\bar X >28.25)=0.9997=99.97\%[/tex]
e) [tex]P(28.25 <\bar X <31.25)=0.9936=99.36%[/tex]
Step-by-step explanation:
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 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".
Let X the random variabl length of advertisements produced by Majesty Video Production Inc. We know from the problem that the distribution for the random variable X is given by:
[tex]X\sim N(\mu =30,\sigma =2)[/tex]
We take a sample of n=16 . That represent the sample size.
a. What can we say about the shape of the distribution of the sample mean time?
From the central limit theorem we know that the distribution for the sample mean [tex]\bar X[/tex] is also normal and is given by:
[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]
[tex]\bar X \sim N(\mu=30, \frac{2}{\sqrt{16}})[/tex]
b. What is the standard error of the mean time?
The standard error is given by this formula:
[tex]Se=\frac{\sigma}{\sqrt{n}}=\frac{2}{\sqrt{16}}=0.5[/tex]
c. What percent of the sample means will be greater than 31.25 seconds?
In order to answer this question we can use the z score in order to find the probabilities, the formula given by:
[tex]z=\frac{\bar X- \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
And we want to find this probability:
[tex]P(\bar X >31.25)=1-P(\bar X<31.25)=1-P(Z<\frac{31.25-30}{0.5})=1-P(Z<2.5)=1-0.994=0.006=0.6\%[/tex]
d. What percent of the sample means will be greater than 28.25 seconds?
In order to answer this question we can use the z score in order to find the probabilities, the formula is given by:
[tex]z=\frac{\bar X- \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
And we want to find this probability:
[tex]P(\bar X >28.25)=1-P(\bar X<28.25)=1-P(Z<\frac{28.25-30}{0.5})=1-P(Z<-3.5)=1-0.00023=0.9997=99.97\%[/tex]
e. What percent of the sample means will be greater than 28.25 but less than 31.25 seconds?"
We want this probability:
[tex]P(28.25 <\bar X <31.25)=P(\frac{28.25-30}{0.5}<Z<\frac{31.25-30}{0.5})=P(-3.5<Z<2.5)=P(Z<2.5)-P(Z<-3.5)=0.9938-0.000233=0.9936=99.36\%[/tex]
- The shape of the distribution of the sample mean time is normal.
- The standard error of the mean time is 0.5.
- The percentage of the sample means that would be greater than 31.25 seconds is 0.621%.
- The percentage of the sample means that would be greater than 28.25 seconds is 99.977%.
- The percentage of the sample means that would be greater than 28.25 but less than 31.25 seconds is 99.357%.
Given the following data:
- Mean = 30.
- Standard deviation = 2.
- Sample = 16 ads.
What is a normal distribution?
A normal distribution is also referred to as the Gaussian distribution and it can be defined as a probability distribution that is continuous and symmetrical on both sides of the mean, which shows that all data near the mean have a higher frequency than data far from the mean.
a. The shape of the distribution of the sample mean time is normal.
b. To calculate the standard error of the mean time, we would apply this formula:
[tex]Standard\;error =\frac{\delta }{\sqrt{n} } \\\\Standard\;error =\frac{2}{\sqrt{16} } \\\\Standard\;error =\frac{2}{4 }[/tex]
Standard error = 0.5.
c. To calculate the percentage of the sample means that would be greater than 31.25 seconds:
[tex]P(\bar{x} > 31.25)=P(Z > 31.25)\\\\P(Z > 31.25)=\frac{31.25-30}{0.5} =2.5\\\\P(Z > 2.5)=1-P(Z < 2.5)\\\\P(Z > 2.5)=1-0.99379\\\\P(Z > 2.5)=0.00621[/tex]
P(Z>2.5) = 0.621%.
d. To calculate the percentage of the sample means that would be greater than 28.25 seconds:
[tex]P(\bar{x} > 28.25)=P(Z > 28.25)\\\\P(Z > 28.25)=\frac{28.25-30}{0.5} =-3.5\\\\P(Z > -3.5)=1-P(Z < -3.5)\\\\P(Z > -3.5)=1-0.00023\\\\P(Z > 2.5)=0.99977[/tex]
P(Z>2.5) = 99.977%.
e. To calculate the percentage of the sample means that would be greater than 28.25 but less than 31.25 seconds:
[tex]P(28.25 < \bar{x} < 31.25)=P(Z > -3.5, Z > 2.5)\\\\P(-3.5 < Z < 2.5)=0.9938-0.00023\\\\P(-3.5 < Z < 2.5)=0.99357[/tex]
P(-3.5<Z<2.5) = 99.357%.
Read more on normal distribution here: https://brainly.com/question/4637344
