Homework SourseThe owner of a fish market has an assistant who has determi
. The owner of a fish market has an assistant who has determined that the weights of catfish are normally distributed, with mean of µ= 3.2 pounds and a standard deviation of 0.8 pounds.
(i)If a sample of 16 fish is taken what is the probability that the sample mean will be between 2.6 and 4.0 pounds? That is , find
P(2.6 less than or equal to MEAN less than or equal to 4.0)
(ii) If a sample of 9 fish is taken what is the probability that the sample mean will be more than 4 pounds? That is , find
P(Mean is greater than 4.0)
(iii) Suppose the population standard deviation is unknown. If 60% of all sample means are greater than 3 pounds and the population mean is still 3.2 pounds, what is the value of the population standard deviation?
Solution
i)
We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as
x1 = lower bound = 2.6
x2 = upper bound = 4
u = mean = 3.2
n = sample size = 16
s = standard deviation = 0.8
Thus, the two z scores are
z1 = lower z score = (x1 - u) * sqrt(n) / s = -3
z2 = upper z score = (x2 - u) * sqrt(n) / s = 4
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.001349898
P(z < z2) = 0.999968329
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.998618431 [ANSWER]
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ii)
We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as
x = critical value = 4
u = mean = 3.2
n = sample size = 9
s = standard deviation = 0.8
Thus,
z = (x - u) * sqrt(n) / s = 3
Thus, using a table/technology, the right tailed area of this is
P(z > 3 ) = 0.001349898 [ANSWER]
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iii)
The z score corresponding to a 0.60 right tailed area, by table/technology, is
z = -0.253347103
Thus, as
z = (x-u)*sqrt(n)/s
Then
-0.253347103 = (3-3.2)*sqrt(9)/s
Then
s = 2.368292325 [ANSWER]
