Question

the reading speed of second grade students is approximately normal, with a mean of 85 words per minute (wpm) and a standard deviation of 11.
a) what is the probability that a single randomly selected student will read more than 91 words per minute?
0.2912
b) what is the probability that a random sample of 20 students will have a mean reading rate of more than 91 words per minute?
n=0
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Explanation:

Part A

Step 1: Identify the distribution and parameters

The reading speed of a single student is normally distributed with mean \(\mu = 85\) wpm and standard deviation \(\sigma = 11\) wpm. We want to find \(P(X>91)\) where \(X\) is the reading speed of a single student.

First, we calculate the z - score using the formula \(z=\frac{x - \mu}{\sigma}\). For \(x = 91\), \(\mu=85\) and \(\sigma = 11\), we have:
\(z=\frac{91 - 85}{11}=\frac{6}{11}\approx0.5455\)

Step 2: Find the probability

We know that \(P(X>91)=1 - P(X\leq91)\). And \(P(X\leq91)=P(Z\leq0.5455)\) (where \(Z\) is the standard normal variable). Looking up the z - score of \(0.5455\) in the standard normal table (or using a calculator with a normal - distribution function), we find that \(P(Z\leq0.5455)\approx0.7088\). Then \(P(X > 91)=1 - 0.7088 = 0.2912\)

Part B

Step 1: Identify the distribution of the sample mean

The sample size \(n = 20\). The sampling distribution of the sample mean \(\bar{X}\) has mean \(\mu_{\bar{X}}=\mu = 85\) and standard deviation \(\sigma_{\bar{X}}=\frac{\sigma}{\sqrt{n}}=\frac{11}{\sqrt{20}}\approx\frac{11}{4.4721}\approx2.46\)

We want to find \(P(\bar{X}>91)\). First, we calculate the z - score for the sample mean using the formula \(z=\frac{\bar{x}-\mu_{\bar{X}}}{\sigma_{\bar{X}}}\). For \(\bar{x}=91\), \(\mu_{\bar{X}} = 85\) and \(\sigma_{\bar{X}}\approx2.46\), we have:
\(z=\frac{91 - 85}{2.46}=\frac{6}{2.46}\approx2.44\)

Step 2: Find the probability

We know that \(P(\bar{X}>91)=1 - P(\bar{X}\leq91)\). And \(P(\bar{X}\leq91)=P(Z\leq2.44)\) (where \(Z\) is the standard normal variable). Looking up the z - score of \(2.44\) in the standard normal table (or using a calculator with a normal - distribution function), we find that \(P(Z\leq2.44)\approx0.9927\). Then \(P(\bar{X}>91)=1 - 0.9927=0.0073\)

Answer:

s:

• Part A: \(0.2912\)
• Part B: \(0.0073\) (the value of \(n = 20\))