1. You will need the same dataset used for problem 3 in homework 1 (the dataset
ID: 3303306 • Letter: 1
Question
1. You will need the same dataset used for problem 3 in homework 1 (the dataset obtained from the yahoo website with your ticker code). Use Excel to calculate the average and standard deviation of the close data column. Assume that these two numbers represent the population (parametric) mean and population standard deviation, respectively, for the variable length (in cm) in a population of a species of fish. Attach a printout of the data to your homework and write down the ticker code on it. 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 [x will be equal to: -(o/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 [x = + (/2)] Calculate the probability of sampling at random a fish that has a size between the two values [x = -(o/2), x = + (/2)] used in parts "a" and "b,” respectively c. Calculate the 25th and 75th percentiles of fish size for the population using the normal distribution table. e. 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: -(o/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]
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