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In: Statistics and Probability

What z-scores bound the middle 94% of the standard normal distribution? Round your answers to two...

  1. What z-scores bound the middle 94% of the standard normal distribution? Round your answers to two decimal places.

Left endpoint:          right endpoint:

  1. For a random variable x which is normally distributed with mean 9.39 and standard deviation 1.7

Q1 =

  1. The number of widgets produced by a particular machine in one day is approximately normally distributed with mean 85.84 and standard deviation 12.95. What is the 32th percentile? (Round your answer to four decimal places.)
  1. Scores on a recent Stat test were normally distributed with mean 71.66 and standard deviation 8.46. What was the lowest score a student could earn and still be in the top 10%? (Round your answer to the nearest integer.)
  1. Until recently, scores of college graduates on the Verbal portion of the GRE test were normally distributed with mean 470 and standard deviation 108. What is the lowest score a student could earn and still be in the 80th percentile? (Round your answer to the nearest integer.)
  1. The heights of plants of a certain species are approximately normally distributed with μ=21.3 cm and σ=5.3 cm. What is the 96th percentile for the heights of these plants? ( Round your answer to four decimal places.)
  1. The pH measurements of water specimens from various locations along a given river basin are normally distributed with mean 8.5 and standard deviation 0.28. Three-quarters of the pH measurements in this river basin are greater than:    (Round your answer to the nearest hundredth.)

Solutions

Expert Solution

Solution(a)
We need to calculate Z-scores bound the middle 94% of the standard normal distribution i.e. P(-z<Z<z) = 0.94
we need to calculate z
Here alpha = 0.94
we need to calculate middle 94% percent so
p-value for upper bound = 0.97
p-value for lower bound = 0.03
From Z table we found Z-score for lower bound = -1.88
From Z table we found Z-score for upper bound = +1.88
So P(-1.88<Z<1.88) = 0.94
Solution(b)
Given in the question
Mean() = 9.39
Standard deviation() = 1.7
We need to calculate Q1 or 25th percentile
So p-value = 0.25, From Z table we found Z-score = -0.6745
So Q1 can be calculated as
Q1 = + Z-score * = 9.39 - 0.6745*1.7 = 8.24
Q1 = 8.24
Solution(c)
Given in the question
Mean() = 85.84
Standard deviation() = 12.95
We need to calculate Q1 or 32th percentile
So p-value = 0.32, From Z table we found Z-score = -0.4677
So 32 percentile can be calculated as
X = + Z-score * = 85.84 - 0.4677*12.95 = 79.7833
32 percentile is 79.7833
Solution(d)
Given in the question
Mean() = 71.66
Standard deviation() = 8.46
We need to calculate the lowest score a student could earn and still be in the top 10%
So p-value = 0.90, From Z table we found Z-score = 1.28155
So Score can be calculated as
X = + Z-score * = 71.66 + 1.28155*8.46 = 82.502 or 83
lowest score is 83 a student could earn and still be in the top 10%
Solution(e)
Given in the question
Mean() = 470
Standard deviation() = 108
We need to calculate the lowest score for 80th percentile
So p-value = 0.80, From Z table we found Z-score = 0.8416
So Score can be calculated as
X = + Z-score * = 470 + 0.8416*108 = 560.89 or 561
80th percentile lowest score = 561
Solution(f)
Mean() = 21.3
Standard deviation() = 5.3
We need to calculate 96th percentile for the heights which can be calculated as
So p-value = 0.96, From Z table we found Z-score = 1.75
So Score can be calculated as
X = + Z-score * = 21.3 + 1.75*5.3 =30.575
96th percentile height = 30.575
Solution(g)
Mean() = 8.5
Standard deviation() = 0.28
We need to calculate 25th percentile so that we can found Three-quarters of the pH measurements in this river basin are greater than
So p-value = 0.25, From Z table we found Z-score = -0.6745
So Score can be calculated as
X = + Z-score * = 8.5 - 0.6745*0.28 =8.31
Three-quarters of the pH measurements in this river basin are greater than 8.31


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