Question

The weight of adobe bricks for construction is normally distributed with a mean of 3 pounds and a standard deviation of 0.25 pound. Assume that the weights of the bricks are independent and that a random sample of 25 bricks is chosen.

a) What is the probability that the mean weight of the sample is less than 3.10 pounds? Round your answer to four decimal places

b) What value will the mean weight exceed with probability 0.99? Round your answer to two decimal places.

Answer 1

Answer:

a) 0.9772 = 97.72% probability that the mean weight of the sample is less than 3.10 pounds

b) 2.88 pounds

Step-by-step explanation:

To solve this question, we have 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:

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

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.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$, a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 3, \sigma = 0.25, n = 25, s = \frac{0.25}{\sqrt{25}} = 0.05$

a) What is the probability that the mean weight of the sample is less than 3.10 pounds? Round your answer to four decimal places

This is the pvalue of Z when X = 3.10. So

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

By the Central Limit Theorem

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

$Z = \frac{3.1 - 3}{0.05}$

$Z = 2$

$Z = 2$ has a pvalue of 0.9772

0.9772 = 97.72% probability that the mean weight of the sample is less than 3.10 pounds

b) What value will the mean weight exceed with probability 0.99? Round your answer to two decimal places.

This is the value of X when Z has a pvalue of 1-0.99 = 0.01. So X when Z = -2.33.

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

$-2.33 = \frac{X - 3}{0.05}$

$X - 3 = -2.33*0.05$

$X = 2.88$