Answer :
Answer:
a) P(12.99 ≤ X ≤ 13.01) = 0.3840
b) P(X ≥ 13.01) = 0.3075
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the cental 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 = 13, \sigma = 0.08[/tex]
(a) Calculate P(12.99 ≤ X ≤ 13.01) when n = 16.
Here we have [tex]n = 16, s = \frac{0.08}{\sqrt{16}} = 0.02[/tex]
This probability is the pvalue of Z when X = 13.01 subtracted by the pvalue of Z when X = 12.99.
X = 13.01
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{13.01 - 13}{0.02}[/tex]
[tex]Z = 0.5[/tex]
[tex]Z = 0.5[/tex] has a pvalue of 0.6915
X = 12.99
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{12.99 - 13}{0.02}[/tex]
[tex]Z = -0.5[/tex]
[tex]Z = -0.5[/tex] has a pvalue of 0.3075
0.6915 - 0.3075 = 0.3840
P(12.99 ≤ X ≤ 13.01) = 0.3840
(b) How likely is it that the sample mean diameter exceeds 13.01 when n = 25?
P(X ≥ 13.01) =
This is 1 subtracted by the pvalue of Z when X = 13.01. So
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{13.01 - 13}{0.02}[/tex]
[tex]Z = 0.5[/tex]
[tex]Z = 0.5[/tex] has a pvalue of 0.6915
1 - 0.6915 = 0.3075
P(X ≥ 13.01) = 0.3075
Answer:
(a) P(12.99 ≤ X ≤ 13.01) = 0.3829
(b) P(X ≥ 13.01) = 0.2660
Step-by-step explanation:
We are given that the inside diameter of a randomly selected piston ring is a random variable with mean value 13 cm and standard deviation 0.08 cm. Suppose the distribution of the diameter is normal.
Firstly, using Central limit theorem the z score probability distribution is given by;
Z = [tex]\frac{X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = mean value = 13 cm
[tex]\sigma[/tex] = standard deviation = 0.08 cm
n = sample size = 16
(a) P(12.99 ≤ X ≤ 13.01) = P(X ≤ 13.01) - P(X < 12.99)
P(X ≤ 13.01) = P( [tex]\frac{X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] ≤ [tex]\frac{13.01-13}{\frac{0.08}{\sqrt{16} } }[/tex] ) = P(Z ≤ 0.50) = 0.69146
P(X < 12.99) = P( [tex]\frac{X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] < [tex]\frac{12.99-13}{\frac{0.08}{\sqrt{16} } }[/tex] ) = P(Z < -0.50) = 1 - P(Z ≤ 0.50)
= 1 - 0.69146 = 0.30854
Therefore, P(12.99 ≤ X ≤ 13.01) = 0.69146 - 0.30854 = 0.3829
(b) Probability that sample mean diameter exceeds 13.01 when n = 25 is given by = P(X ≥ 13.01)
P(X ≥ 13.01) = P( [tex]\frac{X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] ≥ [tex]\frac{13.01-13}{\frac{0.08}{\sqrt{25} } }[/tex] ) = P(Z ≥ 0.625) = 1 - P(Z < 0.625)
= 1 - 0.73401 = 0.2660
Hence, Probability that sample mean diameter exceeds 13.01 when n = 25 is 0.2660.