Respuesta :
Answer:
Answer is 0.1357
Step-by-step explanation:
Let X represent the lifetimes of cell phones.
Given that X is normal with mean = 24.3 and std dev = 2.6
Sample size = 33
Hence std error of sample = 2.6/sq rt 33
=2.6/5.74
=0.453
When x Is normal we have
(x-24.3)/0.453 ~(N(0,1)) i.e. Z variate
P(X<23.8) = P(Z<-0.5/0.453) = P(Z<-1.10)
=0.5-0.3643
=0.1357
Using the normal distribution and the central limit theorem, it is found that there is a 0.1357 = 13.57% probability that the mean lifetime of these phones will be less than 23.8. assume cell phone life is normally distributed variable.
In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- It measures how many standard deviations the measure is from the mean.
- After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
- By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation given by [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
In this problem:
- Mean of 24.3 months, thus [tex]\mu = 24.3[/tex].
- Standard deviation of 2.6 months, thus [tex]\sigma = 2.6[/tex].
- Sample of 33, thus [tex]n = 33, s = \frac{2.6}{\sqrt{33}}[/tex].
The desired probability is the p-value of Z when X = 23.8, thus:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem:
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{23.8 - 24.3}{\frac{2.6}{\sqrt{33}}}[/tex]
[tex]Z = -1.1[/tex]
[tex]Z = -1.1[/tex] has a p-value of 0.1357.
0.1357 = 13.57% probability that the mean lifetime of these phones will be less than 23.8. assume cell phone life is normally distributed variable.
A similar problem is given at https://brainly.com/question/25318038
