Question 438815: Assume that the weights of all packages of a certain brand of cookies are normally distributed with a mean of 26 oz and a standard deviation of .2 ounces. Find the probability that that the mean weight, xbar of a random sample of 50packages of this brand of cookie will be between 25.6 and 26.4 oz.
Mean= 26
St. Dev.= .2
n=50
26.4-26/ .2/sq rt 50= 0.4/0.028284271
25.6-26/ .2/sqrt 50= -0.4/0.028284271
I am getting a postive and negative z value which doesnt exist. Please help= )
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Assume that the weights of all packages of a certain brand of cookies are normally distributed with a mean of 26 oz and a standard deviation of .2 ounces. Find the probability that that the mean weight, xbar of a random sample of 50packages of this brand of cookie will be between 25.6 and 26.4 oz.
Mean= 26
St. Dev.= .2
n=50
(26.4-26)/[0.2/sqrt(50)]= 0.4/0.028284 = 14.14
(25.6-26)/[0.2/sqrt(50(]= -0.4/0.028284 = -14.14
I am getting a positive and negative z value which doesnt exist.
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Negative z-values do exist.
P(25.6<= x-bar <=26.4) = P(-14.14<= z <=14.14) is approximately 1.
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Cheers,
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z values range from -inf to +inf.
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A z value tells you how many standard deviations
a data value is from the mean. A plus z-value
tells you the data value is greater than the mean.
A negative z-value tells you the data value is
less than the mean.
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Cheers,
Stan H.
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