Question

What is the standard error of the mean in this example?

1) 12.6491

What is the likelihood the sample mean is greater than 320 minutes?

2) .7854

What is the likelihood the sample mean is between 320 and 350 minutes?

3) .0569 .7285

Answer 1

According to central limit theorem,

Estimate of population mean, = sample mean = 330 minutes

Sample size, n = 40

Sample standard deviation, = 80

1) Standard error of the mean, s = = 80/sqrt(4)

= 12.6491

2)

For 320 minutes, z value = (320-330)/12.6491 = -0.79057

P(z<-0.79057) = NORMSDIST(-0.79057) = 0.2146

Likelihood that the sample mean is greater than 320 minutes = 1 - 0.2146 = 0.7854

3)

For 350 minutes, z value = (350-330)/12.6491 = 1.5812

P(z<1.5812) = NORMSDIST(1.5812) = 0.9431

Likelihood that the sample mean is between 320 and 350 minutes = P(z<1.5812) - P(z<-079057)

= 0.9431 - 0.2146

= 0.7285