A population has a mean of 300 and a standard deviation of 70. Suppose a sample of size 100 is selected and is used to estimate
1 answer:
Step-by-step explanation:
Sol
Given that ,
mean =300
standard deviation =a =70
n=100
ux=128
a ) within 6 =300 +/-6 : 294, 306
P(294' ' 306)
: P(-0.86 : Z ' 0.86)
: P(Z ' 0.86) - P(Z ' -0.86)
Using z table,
: 0.8051-0.1949
: 0.6102
Probability : 0.6102
a ) within 18 =300 18 : 294, 318
P(294' ' 306)
: P(Z ' 2.57) - P(Z ' -2.57)
Using z table,
:0.9949-0.0051
:0.9898
Probability =0.9898
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Step-by-step explanation:
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Step-by-step explanation: