SOLUTION: QUESTIONS 1-5 ARE BASED ON THE FOLLOWING INFORMATION
The fill in bottles of a wine is normally distributed with mean μ = 750 milliliters (ml) and standard deviation
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Question 456318: QUESTIONS 1-5 ARE BASED ON THE FOLLOWING INFORMATION
The fill in bottles of a wine is normally distributed with mean μ = 750 milliliters (ml) and standard deviation of σ = 9 ml.
1 For samples of size n = 64 bottles selected from this population, the mean of the sampling distribution of x̄ is _____ and the standard error of x̄ ______.
a E(x̄) = 93.75 se(x̄) = 9
b E(x̄) = 750 se(x̄) = 9
c E(x̄) = 750 se(x̄) = 0.141
d E(x̄) = 750 se(x̄) = 1.125
2 "The probability that the mean of a sample of n = 64 bottles falls between 748 and 752 ml is,
P(748 < x̄ < 752) = _____."
a 0.9500
b 0.9250
c 0.8165
d 0.7154
3 The proportion of the means obtained from samples of size n = 64 bottles that fall within ±1 ml from the mean of the population of bottles is _____.
a 0.6827
b 0.6266
c 0.5223
d 0.4039
4 The margin of error for the interval which contains 80% of all sample means is: MOE = _______.
a 2.21 ml.
b 1.85 ml.
c 1.44 ml.
d 1.29 ml.
5 To obtain a margin of error MOE = ± 1 ml for an interval that contains 95% of all sample means, the sample size is n = _____.
a 312
b 217
c 159
d 122
Answer by edjones(8007) (Show Source): You can put this solution on YOUR website!
1)s=9/sqrt(64)=9/8=1.125
m=750
.
2)
z=2/1.125=1.78
area under curve for +-1.78=.9250
.
3)
1/1.125 = .89
area under curve for +-0.89=0.6266
.
4)
For the area under the curve of 0.8 the z=+-1.28
1.28 * 9/8 = 1.44 ml
.
5)
For the area under the curve of 0.95 the z=+-1.96
1.96 * 9/sqrt(x)=1 ml
sqrt(x)=1.96*9=17.64
x=311.17=312 always round up.
.
Ed
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