Question

RADIATION ON MARS

The electronic equipment in the rovers will stop functioning if the rover receives too much radiation. Suppose that the mean amount of radiation received per day on the surface of Mars is 100, and the standard deviation radiation per day is 20. The rovers will be operating on the surface for 600 days. Radiation amounts on different days are independent.

What is the probability that the daily average amount of radiation received by a rover will exceed 102?

Answer 1

0.71% probability that the daily average amount of radiation received by a rover will exceed 102

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 = 100, \sigma = 20, n = 600, s = (20)/(√(600)) = 0.8165`

What is the probability that the daily average amount of radiation received by a rover will exceed 102?

This is 1 subtracted by the pvalue of Z when X = 102. So

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

By the Central Limit Theorem

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

`Z = (102 - 100)/(0.8165)`

`Z = 2.45`

`Z = 2.45` has a pvalue of 0.9929

1 - 0.9929 = 0.0071

0.71% probability that the daily average amount of radiation received by a rover will exceed 102