In a study of facial behavior, people in a control group are timed for eye contact in a 5- minute period. Their times are normally distributed with a mean of 184 seconds and a standard deviation of 55 seconds. For a randomly selected person from the control group, find the probability that the eye contact time is greater than 230 seconds

0.20148

Step-by-step explanation:

We solve using z score formula

z = (x-μ)/σ, where

x is the raw score = 230

μ is the population mean = 184

σ is the population standard deviation = 55

z = 230 - 184/55

z = 0.83636

P-value from Z-Table:

P(x<230) = 0.79852

P(x>230) = 1 - P(x<230) = 0.20148

Therefore, the probability that the eye contact time is greater than 230 seconds is 0.20148

she will sew have sewn five dresses   and be 1/3 done with the sixth one in an hour.

step-by-step explanation:

3/4 of an hour=45 mins

lucy makes 1 dress per 11.25 minutes

45 mins+11.25 mins=56.25 mins

60 mins- 56.25 mins=3.75 mins

3.75 mins is 1/3 of of 11.25 mins

therefore she the other 1/3 would be part of the hour and she will have sewn 5 dresses and begun the sixth.

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