Question

online lectures: the duration of online educational lectures on a learning platform has a mean length of 50.3 minutes and a standard deviation of 12.1 minutes. a sample of 95 lectures is selected. use excel. round the answers to at least four decimal places.
part 1 of 2
(a) find the probability that the mean length of this sample is between 47.5 and 52.
the probability that the mean length of this sample is between 47.5 and 52 is
part 2 of 2
(b) would it be unusual for the mean length to be greater than 53.8? explain.
it be unusual for the mean length to be greater than 53.8 because the probability is , which is the common cutoff of 0.05.

Explanation:

Step1: Identify Distribution

The population mean \(\mu = 50.3\), standard deviation \(\sigma = 12.1\), sample size \(n = 95\). By the Central Limit Theorem, the sampling distribution of the sample mean \(\bar{X}\) is approximately normal with mean \(\mu_{\bar{X}}=\mu = 50.3\) and standard deviation \(\sigma_{\bar{X}}=\frac{\sigma}{\sqrt{n}}=\frac{12.1}{\sqrt{95}}\approx1.2447\).

Step2: Calculate Z - Scores (Part a)

For \(x_1 = 47.5\):
\(z_1=\frac{47.5 - 50.3}{1.2447}=\frac{- 2.8}{1.2447}\approx - 2.25\)
For \(x_2 = 52\):
\(z_2=\frac{52 - 50.3}{1.2447}=\frac{1.7}{1.2447}\approx1.366\)

Step3: Find Probability (Part a)

We need \(P(47.5<\bar{X}<52)=P(-2.25 < Z < 1.366)\)
Using Excel, we can use the formula \(=NORM.DIST(52,50.3,1.2447,TRUE)-NORM.DIST(47.5,50.3,1.2447,TRUE)\)
Calculating \(NORM.DIST(52,50.3,1.2447,TRUE)\approx0.8970\)
Calculating \(NORM.DIST(47.5,50.3,1.2447,TRUE)\approx0.0122\)
So \(P(47.5 < \bar{X}<52)=0.8970 - 0.0122 = 0.8848\)

Step4: Calculate Z - Score (Part b)

For \(x = 53.8\):
\(z=\frac{53.8 - 50.3}{1.2447}=\frac{3.5}{1.2447}\approx2.81\)

Step5: Find Probability (Part b)

We need \(P(\bar{X}>53.8)=1 - P(\bar{X}\leq53.8)\)
Using Excel, \(=1 - NORM.DIST(53.8,50.3,1.2447,TRUE)\)
Calculating \(NORM.DIST(53.8,50.3,1.2447,TRUE)\approx0.9975\)
So \(P(\bar{X}>53.8)=1 - 0.9975 = 0.0025\)
Since \(0.0025<0.05\), it would be unusual.

Answer:

Part (a)

The probability that the mean length of this sample is between 47.5 and 52 is \(\boldsymbol{0.8848}\) (rounded to four decimal places).

Part (b)

It \(\boldsymbol{would}\) be unusual for the mean length to be greater than 53.8 because the probability is \(\boldsymbol{0.0025}\), which is \(\boldsymbol{less}\) than the common cutoff of 0.05.