Respuesta :
Answer:
73.30% probability that the sample mean score will be within 10 of the population mean
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 normally distributed samples can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.
When we are approximating a binomial distribution to a normal one, we have that [tex]\mu = E(X)[/tex], [tex]\sigma = \sqrt{V(X)}[/tex].
Central limit theorem:
The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sample means with size n of at least 30 can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex]
In this problem, we have that:
[tex]\mu = 510, \sigma = 90, n = 100, s = \frac{90}{\sqrt{100}} = 9[/tex]
What is the probability that the sample mean score will be (a) within 10 of the population mean
This is the pvalue of Z when X = 510 + 10 = 520 subtracted by the pvalue of Z when X = 510 - 10 = 500.
X = 520
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{520 - 510}{9}[/tex]
[tex]Z = 1.11[/tex]
[tex]Z = 1.11[/tex] has a pvalue of 0.8665
X = 500
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{500 - 510}{9}[/tex]
[tex]Z = -1.11[/tex]
[tex]Z = -1.11[/tex] has a pvalue of 0.1335
0.8665 - 0.1335 = 0.7330
73.30% probability that the sample mean score will be within 10 of the population mean