Question

The annual rainfall (in inches) in a certain region is normally distributed with mean 43.2 and variance 20.8. Assume rainfall in different years is independent. Find the probability that out of 15 years, at most 2 have rainfall of more than 50 inches.

Answer 1

Answer:

92.24% probability that out of 15 years, at most 2 have rainfall of more than 50 inches.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the binomial probability distribution.

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation(which is the square root of the variance) $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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.

Binomial probability distribution:

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this problem, we have that:

$\mu = 43.2, \sigma = \sqrt{20.8} = 4.56$

Probability that a year has rainfall of more than 50 inches.

pvalue of Z when X = 50. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{50 - 43.2}{4.56}$

$Z = 1.49$

$Z = 1.49$ has a pvalue of 0.9319

1 - 0.9319 = 0.0681

Find the probability that out of 15 years, at most 2 have rainfall of more than 50 inches.

This is $P(X \leq 2)$ when $n = 15, p = 0.0681$. So

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{15,0}.(0.0681)^{0}.(0.9319)^{15} = 0.3472$

$P(X = 1) = C_{15,1}.(0.0681)^{1}.(0.9319)^{14} = 0.3805$

$P(X = 2) = C_{15,2}.(0.0681)^{2}.(0.9319)^{13} = 0.1947$

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.3472 + 0.3805 + 0.1947 = 0.9224$

92.24% probability that out of 15 years, at most 2 have rainfall of more than 50 inches.

Answer 2

Answer:

Probability that out of 15 years, at most 2 have rainfall of more than 50 inches is 0.0443.

Step-by-step explanation:

We are given that the annual rainfall (in inches) in a certain region is normally distributed with mean 43.2 and variance 20.8.

Assume rainfall in different years is independent. We have to find the probability that out of 15 years, at most 2 have rainfall of more than 50 inches.

Firstly, we find the probability of annual rainfall being more than 50 inches in these 15 years.

Let X = annual rainfall (in inches) in a certain region

So, X ~ N($\mu = 43.2, \sigma^{2} = 20.8^{2}$)

The z score probability distribution is given by;

                Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean

            $\sigma$ = standard deviation

So, Probability that annual rainfall is of more than 50 inches is given by = P(X > 50 inches)

      P(X > 50) = P( $\frac{X-\mu}{\sigma}$ > $\frac{50-43.2}{20.8}$ ) = P(Z > 0.33) = 1 - P(Z $\leq$ 0.33)

                                                      = 1 - 0.6293 = 0.3707 or 0.371

Hence, Probability that annual rainfall is of more than 50 inches is 0.371.

Now, we have to find the probability that out of 15 years, at most 2 have rainfall of more than 50 inches.

The above situation can be represented through Binomial distribution;

$P(Y=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; y = 0,1,2,3,.....$

where, n = number of trials (samples) taken = 15 years

            r = number of success = at most 2

            p = probability of success which is of rainfall more than 50 inches,

                   i.e; 0.371.

LET Y = a random variable

So, it means Y ~ $Binom(n=15, p=0.371)$

Now, Probability that out of 15 years, at most 2 have rainfall of more than 50 inches is given by = P(Y $\leq$ 2)

   P(Y $\leq$ 2) = P(Y = 0) + P(Y = 1) + P(Y = 2)

 = $\binom{15}{0}0.371^{0} (1-0.371)^{15-0}+ \binom{15}{1}0.371^{1} (1-0.371)^{15-1}+ \binom{15}{2}0.371^{2} (1-0.371)^{15-2}$

 = $1 \times 1 \times 0.629^{15} +15 \times 0.371^{1} \times 0.629^{14} +105 \times 0.371^{2} \times 0.629^{13}$

 = 0.0443

Therefore, probability that out of 15 years, at most 2 have rainfall of more than 50 inches is 0.0443.