Respuesta :
Answer:
a) 0.6426 = 64.26% probability that the sample mean will be within +/- 5 of the population mean.
b) 0.9544 = 95.44% probability that the sample mean 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
When the distribution is normal, we use 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.
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.
In this question, we have that:
[tex]\mu = 200, \sigma = 50, n = 100, s = \frac{50}{\sqrt{100}} = 5[/tex]
a. What is the probability that the sample mean will be within +/- 5 of the population mean (to 4 decimals)?
This is the pvalue of Z when X = 200 + 5 = 205 subtracted by the pvalue of Z when X = 200 - 5 = 195.
Due to the Central Limit Theorem, Z is:
[tex]Z = \frac{X - \mu}{s}[/tex]
X = 205
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{205 - 200}{5}[/tex]
[tex]Z = 1[/tex]
[tex]Z = 1[/tex] has a pvalue of 0.8413.
X = 195
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{195 - 200}{5}[/tex]
[tex]Z = -1[/tex]
[tex]Z = -1[/tex] has a pvalue of 0.1587.
0.8413 - 0.1587 = 0.6426
0.6426 = 64.26% probability that the sample mean will be within +/- 5 of the population mean.
b. What is the probability that the sample mean will be within +/- 10 of the population mean (to 4 decimals)?
This is the pvalue of Z when X = 210 subtracted by the pvalue of Z when X = 190.
X = 210
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{210 - 200}{5}[/tex]
[tex]Z = 2[/tex]
[tex]Z = 2[/tex] has a pvalue of 0.9772.
X = 195
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{190 - 200}{5}[/tex]
[tex]Z = -2[/tex]
[tex]Z = -2[/tex] has a pvalue of 0.0228.
0.9772 - 0.0228 = 0.9544
0.9544 = 95.44% probability that the sample mean will be within +/- 10 of the population mean.
(a): The required probability is [tex]P(195 < \bar{x} < 205)=0.6826[/tex]
(b): The required probability is [tex]P(190 < \bar{x} < 200)=0.9544[/tex]
Z-score:
A numerical measurement that describes a value's relationship to the mean of a group of values.
Given that,
mean=200
Standard deviation=50
[tex]n=100[/tex]
[tex]\mu_{\bar{x}}=200[/tex]
[tex]\sigma{\bar{x}} =\frac{\sigma}{\sqrt{n} } \\=\frac{50}{\sqrt{100} }\\ =5[/tex]
Part(a):
within [tex]5=200\pm 5=195,205[/tex]
[tex]P(195 < \bar{x} < 205)=P(-1 < z < 1)\\=P(z < 1)-P(z < -1)\\=0.8413-0.1587\\=0.6826[/tex]
Part(b):
within [tex]10=200\pm 10=190,200[/tex]
[tex]P(190 < \bar{x} < 200)=P(-1 .98 < z < 1.98)\\=P(z < 2)-P(z < -2)\\=0.9772-0.0228\\=0.9544[/tex]
Learn more about the topic Z-score:
https://brainly.com/question/5512053
