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hram777 [196]
3 years ago
8

The time required to finish a test in normally distributed with a mean of 80 minutes and a standard deviation of 15 minutes. Wha

t is the probability that a student chosen at random will finish the test in more than 95 minutes?
Mathematics
1 answer:
saul85 [17]3 years ago
6 0
The z-score for 95 minutes is:
z=\frac{X-\mu}{\sigma}=\frac{95-80}{15}=1
Reference to a standard normal distribution shows that the required probability is approximately 0.16 or 16%.
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Read 2 more answers
Records show that Oliver is typically 10-30 minutes late
Ratling [72]

The percentage of the time will Oliver get a written warning when  he is more than 16 minutes late for his shift at work is 46.7%.

<h3>What is a density curve?</h3>

The density curve is a type or way of representation of numbered distribution data over a graph. Here, the outcomes are continues. The density curve can not be below the x-axis or horizontal line of the graph.

In a density curve,

  • The mean of a density curve is indicated by μ (mu).
  • The standard deviation of a density curve is indicated by Sx.

Records show that Oliver is typically 10-30 minutes late for his shift at work. The distribution for the minutes he is late forms a consistent pattern, which can be graphed as the given uniform density curve.

Oliver will get a written warning if he is more than 16 minutes late for work. The time remain after the 16 minutes is,

t=30-16

t=14 min

Thus, the percentage of the time will Oliver get a written warning is,

p=(14/30)*100

p=46.7 %

Thus, the percentage of the time will Oliver get a written warning when  he is more than 16 minutes late for his shift at work is 46.7%.

Learn more about the density curve here;

brainly.com/question/25334760

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