How many close friends do you have? Suppose that the number of close friends adults claim to have varies from person to person with mean = 5 and standard deviation = 1.8.
An opinion poll asks this question of an SRS of 900 adults.
We will see later that in this situation the sample mean response has approximately the Normal distribution with mean 5 and standard deviation 0.060.
What is , the probability that the sample result estimates the population truth = 5 to within 0.2?
An opinion poll asks this question of an SRS of 900 adults.
We will see later that in this situation the sample mean response has approximately the Normal distribution with mean 5 and standard deviation 0.060.
What is , the probability that the sample result estimates the population truth = 5 to within 0.2?
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Let Xbar represent the sample mean.
Xbar has approximately the normal distribution with mean 5 and standard deviation 0.060 .
P(Xbar is within 0.2 of 5)
= P(4.8 <= Xbar <= 5.2)
= P((4.8 - 5)/0.06 <= Z <= (5.2 - 5)/0.06)
= P(-3.33 <= Z <= 3.33)
= P(Z <= 3.33) - P(Z < -3.33)
= P(Z <= 3.33) - P(Z > 3.33) by symmetry of the normal curve
= P(Z <= 3.33) - [1 - P(Z <= 3.33)]
= 2P(Z <= 3.33) - 1
= 2(0.9996) - 1 from the normal table
= 0.9992 .
Lord bless you today!
Xbar has approximately the normal distribution with mean 5 and standard deviation 0.060 .
P(Xbar is within 0.2 of 5)
= P(4.8 <= Xbar <= 5.2)
= P((4.8 - 5)/0.06 <= Z <= (5.2 - 5)/0.06)
= P(-3.33 <= Z <= 3.33)
= P(Z <= 3.33) - P(Z < -3.33)
= P(Z <= 3.33) - P(Z > 3.33) by symmetry of the normal curve
= P(Z <= 3.33) - [1 - P(Z <= 3.33)]
= 2P(Z <= 3.33) - 1
= 2(0.9996) - 1 from the normal table
= 0.9992 .
Lord bless you today!