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Sagot :
Answer:
a) By the Central Limit Theorem, it has an approximately normal shape, with mean(center) 0.9 and standard deviation(spread) 0.035.
b) Average numbers of children between 0.83 and 0.97 are reasonably likely in the sample.
c) 0.0021 = 0.21% probability that the average number of children per family in the sample will be 0.8 or less
d) 0.9958 = 99.58% probability that the average number of children per family in the sample will be between 0.8 and 1.0
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal Probability Distribution
Problems of normal distributions can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
Central Limit Theorem
The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
Mean 0.9 and standard deviation 1.1.
This means that [tex]\mu = 0.9, \sigma = 1.1[/tex]
Suppose a television network selects a random sample of 1000 families in the United States for a survey on TV viewing habits.
This means that [tex]n = 1000, s = \frac{1.1}{\sqrt{1000}} = 0.035[/tex]
a. Describe (as shape, center and spread) the sampling distribution of the possible values of the average number of children per family.
By the Central Limit Theorem, it has an approximately normal shape, with mean(center) 0.9 and standard deviation(spread) 0.035.
b. What average numbers of children are reasonably likely in the sample?
By the Empirical Rule, 95% of the sample is within 2 standard deviations of the mean, so:
0.9 - 2*0.035 = 0.83
0.9 + 2*0.035 = 0.97
Average numbers of children between 0.83 and 0.97 are reasonably likely in the sample.
c. What is the probability that the average number of children per family in the sample will be 0.8 or less?
This is the p-value of Z when X = 0.8. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{0.8 - 0.9}{0.035}[/tex]
[tex]Z = -2.86[/tex]
[tex]Z = -2.86[/tex] has a p-value of 0.0021
0.0021 = 0.21% probability that the average number of children per family in the sample will be 0.8 or less.
d. What is the probability that the average number of children per family in the sample will be between 0.8 and 1.0?
p-value of Z when X = 1 subtracted by the p-value of Z when X = 0.8.
X = 1
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{1 - 0.9}{0.035}[/tex]
[tex]Z = 2.86[/tex]
[tex]Z = 2.86[/tex] has a p-value of 0.9979
X = 0.8
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{0.8 - 0.9}{0.035}[/tex]
[tex]Z = -2.86[/tex]
[tex]Z = -2.86[/tex] has a p-value of 0.0021
0.9979 - 0.0021 = 0.9958
0.9958 = 99.58% probability that the average number of children per family in the sample will be between 0.8 and 1.0
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