Answer:
0.04746
Step-by-step explanation:
To answer this one needs to find the area under the standard normal curve to the left of 5 minutes when the mean is 4 minutes and the std. dev. is 0.6 minutes. Either use a table of z-scores or a calculator with probability distribution functions.
In this case I will use my old Texas Instruments TI-83 calculator. I select the normalcdf( function and type in the following arguments: :
normalcdf(-100, 5, 4, 0.6). The result is 0.952. This is the area under the curve to the left of x = 5. But we are interested in finding the probability that a conversation lasts longer than 5 minutes. To find this, subtract 0.952 from 1.000: 0.048. This is the area under the curve to the RIGHT of x = 5.
This 0.048 is closest to the first answer choice: 0.04746.
Answer:
ur answer
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Answer:
Step-by-step explanation:
32 64 128 are the answers
Answer:
Step-by-step explanation:
Since the results for the standardized test are normally distributed, we would apply the formula for normal distribution which is expressed as
z = (x - µ)/σ
Where
x = test reults
µ = mean score
σ = standard deviation
From the information given,
µ = 1700 points
σ = 75 points
We want to the probability that a student will score more than 1700 points. This is expressed as
P(x > 1700) = 1 - P(x ≤ 1700)
For x = 1700,
z = (1700 - 1700)/75 = 0/75 = 0
Looking at the normal distribution table, the probability corresponding to the z score is 0.5
P(x > 1700) = 1 - 0.5 = 0.5
