Respuesta :
Answer:
a) P(X < 14) = 0.0228.
b) 49 batteries must be tested.
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal probability distribution
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
In this problem, we have that:
[tex]\mu = 15, \sigma = 3, n = 36, s = \frac{3}{\sqrt{36}} = 0.5[/tex]
(a) If the claim is true, what is P(X < 14)?
This is the pvalue of Z when X = 14. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{14 - 15}{0.5}[/tex]
[tex]Z = -2[/tex]
[tex]Z = -2[/tex] has a pvalue of 0.0228
So
P(X < 14) = 0.0228.
(b) How many batteries must be tested so that the P(X < 14) = 0.01?
Now we want to find the size of the sample for which X = 14 is in the first percentile, that is, Z when X = 14 has a pvalue of 0.01. So it is Z = -2.327.
First we have to find the standard deviation of the sample in this case.
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]-2.327 = \frac{14 - 15}{s}[/tex]
[tex]-2.327s = -1[/tex]
[tex]2.327s = 1[/tex]
[tex]s = \frac{1}{2.327}[/tex]
[tex]s = 0.43[/tex]
We know that
[tex]s = \frac{\sigma}{\sqrt{n}}[/tex]
So
[tex]0.43 = \frac{3}{\sqrt{n}}[/tex]
[tex]0.43\sqrt{n} = 3[/tex]
[tex]\sqrt{n} = \frac{3}{0.43}[/tex]
[tex](\sqrt{n})^{2} = (\frac{3}{0.43})^{2}[/tex]
[tex]n = 48.7[/tex]
Rounding up
49 batteries must be tested.
