Anne has 20 jobs that she must do in sequence, with the times required to do each of these jobs being independent random variables with mean 50 minutes and standard deviation 10 minutes. mary has 20 jobs that she must do in sequence, with the times required to do each of these jobs being independent random variables with mean 52 minutes and standard deviation 15 minutes. what is the probability that anne finishes before mary?

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victoriadorvilu

21.03.2023

Mathematics

a. Mean assembly time is estimated to fall between 15.6mins and 16.8mins with a 92% CI.

b.636 workers

Step-by-step explanation:

Let $\mu$ denote the mean assembly time in minutes.

Standard deviation, $\sigma=3.6min$

To obtain a 92%CI for $\mu$ based on given parameters: n=120, $\bar X=16.2min$ , $\alpha =1-0.92=0.08,\sigma=3.6$ . $\sigma$ is known, therefore we use the interval:

$\bar X\pm z_0_._5_n\frac{\sigma}{\sqrt n}=16.2\pm 1.75\frac{3.6}{\sqrt 120}=16.2\pm0.575\\=[15.6,16.8]$

Where z_0_._0_4=1.75 , and is obtained from the z-interval tables.

b. A sample of size n needs to be completed based on the info: $\alpha =0.08, \sigma=3.6, E=15sec=0.25min$ , E denotes the margin of error.

Therefore the sample size is calculated as:-

$n=(\frac{z_\frac{\alpha}{z^\sigma}}{E})^2=(\frac{1.75\times3.6}{0.25})^2\\=636$

Hence, 636 workers should be involved in this study in order to have the mean assembly time required.

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