The length of Atlantic herring is normally distributed with a mean of 30.0 centi
ID: 3160339 • Letter: T
Question
The length of Atlantic herring is normally distributed with a mean of 30.0 centimeters, and a standard deviation of 7.0 centimeters. a) what percent of Atlantic herring are shorter than 25.0 centimeters b) what is the 22st percentile of the lengths of Atlantic herring? 2) A random sample of 20 Idaho potatoes has a sample mean weight of 154 grams per potato, and a sample standard deviation of 15 grams per potato. Assume that the weights of Idaho potatoes is normally distributed. a) Give a 95% confidence interval for the mean weight of an Idaho potato:Explanation / Answer
1.
a)
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 25
u = mean = 30
s = standard deviation = 7
Thus,
z = (x - u) / s = -0.714285714
Thus, using a table/technology, the left tailed area of this is
P(z < -0.714285714 ) = 0.237525262 = 23.7525262% [ANSWER]
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b)
First, we get the z score from the given left tailed area. As
Left tailed area = 0.22
Then, using table or technology,
z = -0.772193214
As x = u + z * s,
where
u = mean = 30
z = the critical z score = -0.772193214
s = standard deviation = 7
Then
x = critical value = 24.5946475 [ANSWER]
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