## Without doing any computation, put the following in order from least to greatest, assuming the population is normally distributed with μ=200 and σ=15.(a) P(190≤x≤210) for a random sample of size n=10(b) P(190≤x≤210) for a random sample of size n=20(c) P(190≤x≤210)

### To put the following in order from least to greatest, we need to consider the properties of the normal distribution

To put the following in order from least to greatest, we need to consider the properties of the normal distribution.

(a) P(190≤x≤210) for a random sample of size n=10:

When calculating probabilities for sample means, we use the Central Limit Theorem, which states that for a random sample of size n from any population (regardless of its shape), the distribution of the sample means approaches a normal distribution as n increases.

Since the population mean is μ = 200 and the standard deviation is σ = 15, for a sample size of n = 10, the standard deviation of the sample means is σ/√n = 15/√10.

To calculate P(190≤x≤210) for a random sample of size n=10, we need to find the z-scores corresponding to the lower and upper bounds of the interval.

For 190:

z = (190 – μ) / (σ/√n)

z = (190 – 200) / (15/√10)

z = -10 / (15/√10)

z = -10 / (15/√10)

z = -2√10 / 3

For 210:

z = (210 – μ) / (σ/√n)

z = (210 – 200) / (15/√10)

z = 10 / (15/√10)

z = 10 / (15/√10)

z = 2√10 / 3

Using a standard normal distribution table or calculator, we can find the probabilities corresponding to these z-scores and subtract them to find P(190≤x≤210).

P(190≤x≤210) for a random sample of size n=10 = P(z ≤ 2√10 / 3) – P(z ≤ -2√10 / 3)

(b) P(190≤x≤210) for a random sample of size n=20:

For a sample size of n = 20, the standard deviation of the sample means is σ/√n = 15/√20.

Following the same steps as before, we can find P(190≤x≤210) for a random sample of size n=20 using z-scores.

P(190≤x≤210) for a random sample of size n=20 = P(z ≤ 2√10 / (3√2)) – P(z ≤ -2√10 / (3√2))

(c) P(190≤x≤210) for a random sample of size n:

The standard deviation of the sample means decreases as the sample size increases. As the sample size approaches infinity, the standard deviation of the sample means approaches 0.

For very large sample sizes, the probability P(190≤x≤210) for a random sample will approach 1, meaning that it becomes almost certain that the sample mean falls within the given range.

Therefore, the order from least to greatest is:

(c) P(190≤x≤210) for a random sample of size n >>

(b) P(190≤x≤210) for a random sample of size n = 20 >>

(a) P(190≤x≤210) for a random sample of size n = 10

##### More Answers:

Comparing Exam Scores | ACT vs. SAT – Determining Your Performance Relative to Mean and Standard DeviationApproximating Binomial Probability Distribution Using Normal Distribution | Finding Probability of Exactly 12 Defective Parts in a Shipment

Calculating Probability of More than 684 Americans Supporting the Bill Using Normal Distribution Approximation