Respuesta :
Answer:
96.08% probability that the average length of a randomly selected bundle of steel rods is less than 178.3-cm.
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 = 177.5, \sigma = 1.2, n = 7, s = \frac{1.2}{\sqrt{7}} = 0.4536[/tex]
Find the probability that the average length of a randomly selected bundle of steel rods is less than 178.3-cm.
This is the pvalue of Z when X = 178.3. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{178.3 - 177.5}{0.4536}[/tex]
[tex]Z = 1.76[/tex]
[tex]Z = 1.76[/tex] has a pvalue of 0.9608
96.08% probability that the average length of a randomly selected bundle of steel rods is less than 178.3-cm.
Answer: the probability that the average length of a randomly selected bundle of steel rods is less than 178.3-cm is 0.96
Step-by-step explanation:
Since the lengths of the steel rods are normally distributed, then according to the central limit theorem,
z = (x - µ)/(σ/√n)
Where
x = sample mean lengths of the steel rods
µ = population mean length of the steel rods
σ = standard deviation
n = number of samples
From the information given,
µ = 177.5 cm
x = 178.3 cm
σ = 1.2 cm
n = 7
the probability that the average length of a randomly selected bundle of steel rods is less than 178.3-cm is expressed as
P(x < 178.3)
Therefore,
z = (178.3 - 177.5)/(1.2/√7) = 1.76
Looking at the normal distribution table, the probability corresponding to the z score is 0.96
Therefore,
P(x < 178.3) = 0.96
