Question

Suppose that the weights of 1-year-old boys are approximately normally distributed with a mean of 22.8 lbs and a standard deviation of about 2.15 lbs. Part 1: Find the probability that a randomly selected 1-year old boy from the population will have a weight that is less than 25 lbs. Part 2: Find the probability that the mean weight for a sample of size 10 1-year-old boys will be less than 25 lbs. Part 3: Explain the difference between Part 1 and Part 2.

Answer 1

1: 0.8461 = 84.61% probability that a randomly selected 1-year old boy from the population will have a weight that is less than 25 lbs.

2: 0.9994 = 99.94% probability that the mean weight for a sample of size 10 1-year-old boys will be less than 25 lbs

3: In part 2, we use the sampling distribution of the sample means, which has more values closer to the mean due to the smaller standard error, so a higher probability of finding a mean less than 25 lbs.

Explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the z-score 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 p-value, 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.

Mean of 22.8 lbs and a standard deviation of about 2.15 lbs.

This means that
`\mu = 22.8, \sigma = 2.15`

Part 1: Find the probability that a randomly selected 1-year old boy from the population will have a weight that is less than 25 lbs.

This is the pvalue of Z when X = 25. So

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

`Z = (25 - 22.8)/(2.15)`

`Z = 1.02`

`Z = 1.02` has a pvalue of 0.8461

0.8461 = 84.61% probability that a randomly selected 1-year old boy from the population will have a weight that is less than 25 lbs.

Part 2: Find the probability that the mean weight for a sample of size 10:

Now
`n = 10, s = (2.15)/(√(10))`. So

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

By the Central Limit Theorem

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

`Z = (25 - 22.8)/((2.15)/(√(10)))`

`Z = 3.24`

`Z = 3.24` has a pvalue of 0.9994

0.9994 = 99.94% probability that the mean weight for a sample of size 10 1-year-old boys will be less than 25 lbs

Part 3: Explain the difference between Part 1 and Part 2.

In part 2, we use the sampling distribution of the sample means, which has more values closer to the mean due to the smaller standard error, so a higher probability of finding a mean less than 25 lbs.