Question

What is the probability that a random sample of 10 second grade students from the city results in a mean reading rate of more than 96 words per​ minute

Answer 1

Using the normal distribution, it is found that there is a 0.1029 = 10.29% probability that the sample mean is above 96.

Normal Probability Distribution

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

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

• The z-score measures how many standard deviations the measure is above or below the mean.

• Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

• By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation $s = \frac{\sigma}{\sqrt{n}}$.

Researching the problem on the internet, the parameters are given as follows:

$\mu = 92, \sigma = 10, n = 10, s = \frac{10}{\sqrt{10}} = 3.1623$

The probability that the sample mean is above 96 is one subtracted by the p-value of Z when X = 96, hence:

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

By the Central Limit Theorem:

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

$Z = \frac{96 - 92}{3.1623}$

Z = 1.265

Z = 1.265 has a p-value of 0.8971.

1 - 0.8971 = 0.1029 = 10.29% probability that the sample mean is above 96.

More can be learned about the normal distribution at brainly.com/question/4079902

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