A computer tallied the time to work for 200 days and found it reasonable to normal curve. The main Thomas 35 minutes, and the standard deviation with six minutes. For the 200 workday experiment, find the percent of tonic me to work more than 41 minutes.

We are given that a computer tallied the time to work for 200 days and found it reasonable to the normal curve. The mean is 35 minutes, and the standard deviation with six minutes.

Let X = the time taken by computer to work for 200 days.

So, X ~ Normal([tex]\mu=35, \sigma^{2} =6^{2}[/tex])

The z-score probability distribution for the normal distribution is given by;

Z = [tex]\frac{X-\mu}{\sigma }[/tex] ~ N(0,1)

where, [tex]\mu[/tex] = population mean time = 35 minutes

[tex]\sigma[/tex] = standard deviation = 6 minutes

Now, the probability that computers work more than 41 minutes is given by = P(X > 41 minutes)

P(X > 41 minutes) = P( [tex]\frac{X-\mu}{\sigma }[/tex] > [tex]\frac{41-35}{6 }[/tex] ) = P(Z > 1) = 1 - P(Z [tex]\leq[/tex] 1)

= 1 - 0.84134 = 0.15866

The above probability is calculated by looking at the value of x = 1 in the z table which has an area of 0.84134.

A computer tallied the time to work for 200 days and found it reasonable to normal curve. The main Thomas 35 minutes, and the standard deviation with six minutes. For the 200 workday experiment, find the percent of tonic me to work more than 41 minutes.￼

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A computer tallied the time to work for 200 days and found it reasonable to normal curve. The main Thomas 35 minutes, and the standard deviation with six minutes. For the 200 workday experiment, find the percent of tonic me to work more than 41 minutes.