Question

The mean weight of an adult is 60 kilograms with a variance of 100. If 118 adults are randomly selected, what is the probability that the sample mean would be greater than 62.2 kilograms? Round your answer to four decimal places.

Answer 1

To find the probability that the sample mean would be greater than 62.2 kilograms in a sample of 118 adults with a mean weight of 60 kilograms and a variance of 100, we calculate the z-score and find the area under the standard normal distribution curve to the right of that z-score. The probability is approximately 0.0014 (or 0.14%).

Step-by-step explanation:

To find the probability that the sample mean would be greater than 62.2 kilograms, we need to convert the sample mean into a z-score and then find the area under the standard normal distribution curve to the right of that z-score.

First, we need to calculate the standard deviation of the sample mean using the formula: standard deviation of the sample mean = standard deviation of the population / square root of sample size. In this case, the standard deviation of the population is the square root of the variance, which is 10. Therefore, the standard deviation of the sample mean is 10 / square root of 118, which is approximately 0.917.

Next, we calculate the z-score by subtracting the population mean from the sample mean and then dividing by the standard deviation of the sample mean. So, the z-score is (62.2 - 60) / 0.917, which is approximately 2.986.

Finally, we find the area under the standard normal distribution curve to the right of the z-score using a standard normal distribution table or a calculator. The probability is equal to 1 minus the cumulative probability to the left of the z-score. In this case, the probability is approximately 0.0014 (or 0.14%).

Answer 2

Step 1

Given; The mean weight of an adult is 60 kilograms with a variance of 100. If 118 adults are randomly selected, what is the probability that the sample mean would be greater than 62.2 kilograms? Round your answer to four.

Step 2

In a set with mean and standard deviation, 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.

The Central Limit Theorem establishes that, for a random variable X, with mean and standard deviation, the sample means with size n of at least 30 can be approximated to a normal distribution with mean and standard deviation;

`s=(\sigma)/(√(n))`
`\begin{gathered} \mu=60kg \\ s=√(variance)=√(100)=10 \\ n=118 \\ \end{gathered}`
`\begin{gathered} s=(10)/(√(118)) \\ s=(5√(118))/(59) \end{gathered}`

What is the probability that the sample mean would be greater than 62.2 kilograms?

`\begin{gathered} z=(62.2-60)/((5√(118))/(59)) \\ z=2.38981 \end{gathered}`
`The\text{ p-value=0.0084285362}`

Thus;

`0.0084284970`

Answer;

`0.0084\text{ to 4 d.p}`