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Sagot :
Answer:
0.0062 = 0.62% probability that the sample mean is less than $650.
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 estabilishes 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 of $675 and a standard deviation of $80.
This means that [tex]\mu = 675, \sigma = 80[/tex]
A random sample of 64 families in this city paying for childcare is selected.
This means that [tex]n = 64, s = \frac{80}{\sqrt{64}} = 10[/tex]
Find the probability that the sample mean is less than $650.
This is the pvalue of Z when X = 650.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{650 - 675}{10}[/tex]
[tex]Z = -2.5[/tex]
[tex]Z = -2.5[/tex] has a pvalue of 0.0062
0.0062 = 0.62% probability that the sample mean is less than $650.
The probability that the sample mean is less than $650 is 0.62%.
The z score is given by:
[tex]z=\frac{x-\mu}{\sigma/\sqrt{n} } \\\\where\ x=raw\ score,\sigma=standard\ deviation,\mu=mean,n=sample\ size\\\\\\Given \ \mu=675,\sigma=80,n=84, hence:\\\\For\ x<650:\\\\z=\frac{650-675}{80/\sqrt{64} } =-2.5[/tex]
From the normal distribution table:
P(x < 650) = P(z < -2.5) = 0.0062 = 0.62%
The probability that the sample mean is less than $650 is 0.62%.
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