Please show work (step by step) and calculation. Assume 3.03 and 0.238 represent
ID: 3126639 • Letter: P
Question
Please show work (step by step) and calculation.
Assume 3.03 and 0.238 represent the population (=parametric) mean and population standard deviation, respectively, for the variable length (in cm) in a population of a species of fish.
a) Calculate the probability of sampling at random a fish that is smaller in size than the value you would obtain by subtracting half the standard deviation from the average [ – (/2)]
b) Calculate the probability of sampling at random a fish that is greater in size than the value you would obtain by adding half the standard deviation from the average [ + (/2)]
c) Calculate the probability of sampling at random a fish that has a size between the two values [ – (/2), + (/2)] used in parts “a” and “b,” respectively
d) Calculate the 25th and 75th percentiles of fish size for the population.
Imagine that 5 individuals are sampled at random from this fish population. Calculate the probability that the average calculated will be less than the value: – (/3)
NOTE: Assume the variable is normally distributed and use bell-shaped curve diagrams to shade the areas that correspond with the answers to questions “a” through “d”
Explanation / Answer
a)
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 3.03 - 0.238/2 = 2.911
u = mean = 3.03
s = standard deviation = 0.238
Thus,
z = (x - u) / s = -0.5
Thus, using a table/technology, the left tailed area of this is
P(z < -0.5 ) = 0.308537539 [ANSWER]
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b)
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 3.03+0.238/2 = 3.149
u = mean = 3.03
s = standard deviation = 0.238
Thus,
z = (x - u) / s = 0.5
Thus, using a table/technology, the right tailed area of this is
P(z > 0.5 ) = 0.308537539 [ANSWER]
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C)
We first get the z score for the two values. As z = (x - u) / s, then as
x1 = lower bound = 2.911
x2 = upper bound = 3.149
u = mean = 3.03
s = standard deviation = 0.238
Thus, the two z scores are
z1 = lower z score = (x1 - u)/s = -0.5
z2 = upper z score = (x2 - u) / s = 0.5
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.308537539
P(z < z2) = 0.691462461
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.382924923 [ANSWER]
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d)
First, we get the z score from the given left tailed area. As
Left tailed area = 0.25
Then, using table or technology,
z = -0.67448975
As x = u + z * s,
where
u = mean = 3.03
z = the critical z score = -0.67448975
s = standard deviation = 0.238
Then
x = critical value = 2.869471439 [ANSWER, 25TH PERCENTILE]
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First, we get the z score from the given left tailed area. As
Left tailed area = 0.75
Then, using table or technology,
z = 0.67448975
As x = u + z * s,
where
u = mean = 3.03
z = the critical z score = 0.67448975
s = standard deviation = 0.238
Then
x = critical value = 3.190528561 [ANSWER, 75TH PERCENTILE]
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e)
We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as
x = critical value = 3.03 - 0.238/3 = 2.950666667
u = mean = 3.03
n = sample size = 5
s = standard deviation = 0.238
Thus,
z = (x - u) * sqrt(n) / s = -0.745355992
Thus, using a table/technology, the left tailed area of this is
P(z < -0.745355992 ) = 0.22802827 [ANSWER]