A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 260.5-cm and a standard deviation of 1.6-cm. For shipment, 8 steel rods are bundled together. Find the probability that the average length of a randomly selected bundle of steel rods is greater than 260.2-cm. P(M > 260.2-cm) = Enter your answer as a number accurate to 4 decimal places. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.

P(M > 260.2-cm) = 0.702

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 and standard deviation , the zscore of a measure X is given by:

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 and standard deviation , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean and standard deviation .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

P(M > 260.2-cm)

This is 1 subtracted by the pvalue of Z when X = 260.2. So

By the Central Limit Theorem

has a pvalue of 0.298.

1 - 0.298 = 0.702

So

P(M > 260.2-cm) = 0.702

you can first factor out the n to get n(1+5). we know that 1+5 is 6, so we just have 6n.