Question

Suppose that the US plans to send a shipment of "rovers" to Mars. These are mobile robots, programmed to collect rock and soil samples, and then return to the landing site. The rovers operate independently of each other. The mean weight a rover is programmed to collect is 50 pounds, and the standard deviation of weights is 5 pounds. Weights collected by rovers are approximately normally distributed. If the US sends 10 rovers, what is the probability that the average weight of rock and soil brought back by these rovers will be between 48 pounds and 52 pounds? What sampling distribution should we use to compute this probability?

Answer 1

79.24% probability that the average weight of rock and soil brought back by these rovers will be between 48 pounds and 52 pounds

We use the sampling distribution of the sample means of size 10 to solve this question, by the Central Limit Theorem. They are normally distributed with mean 50 and standard deviation 1.58.

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
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

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
`\mu` and standard deviation
`\sigma`, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
`\mu` and standard deviation
`s = (\sigma)/(√(n))`.

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:

`\mu = 50, \sigma = 5, n = 10, s = (5)/(√(10)) = 1.58`

What is the probability that the average weight of rock and soil brought back by these rovers will be between 48 pounds and 52 pounds?

This is the pvalue of Z when X = 52 subtracted by the pvalue of Z when X = 48. So

X = 52

`Z = (X - \mu)/(\sigma)`

By the Central Limit Theorem

`Z = (X - \mu)/(s)`

`Z = (52 - 50)/(1.58)`

`Z = 1.26`

`Z = 1.26` has a pvalue of 0.8962

X = 48

`Z = (X - \mu)/(s)`

`Z = (48 - 50)/(1.58)`

`Z = -1.26`

`Z = -1.26` has a pvalue of 0.1038

0.8962 - 0.1038 = 0.7924

79.24% probability that the average weight of rock and soil brought back by these rovers will be between 48 pounds and 52 pounds

What sampling distribution should we use to compute this probability?

We use the sampling distribution of the sample means of size 10 to solve this question, by the Central Limit Theorem. They are normally distributed with mean 50 and standard deviation 1.58.