Question

A population has a mean of 200 and a standard deviation of 50. Suppose a sample of size 100 is selected and x is used to estimate μ. (Round your answers to four decimal places.)

Required:
a. What is the probability that the sample mean will be within +/- 5 of the population mean (to 4 decimals)?
b. What is the probability that the sample mean will be within +/- 10 of the population mean (to 4 decimals)?

Answer 1

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 $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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 $\mu$ and standard deviation $\sigma$, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$.

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:

$\mu = 200, \sigma = 50, n = 100, s = \frac{50}{\sqrt{100}} = 5$

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:

$Z = \frac{X - \mu}{s}$

X = 205

$Z = \frac{X - \mu}{s}$

$Z = \frac{205 - 200}{5}$

$Z = 1$

$Z = 1$ has a pvalue of 0.8413.

X = 195

$Z = \frac{X - \mu}{s}$

$Z = \frac{195 - 200}{5}$

$Z = -1$

$Z = -1$ 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

$Z = \frac{X - \mu}{s}$

$Z = \frac{210 - 200}{5}$

$Z = 2$

$Z = 2$ has a pvalue of 0.9772.

X = 195

$Z = \frac{X - \mu}{s}$

$Z = \frac{190 - 200}{5}$

$Z = -2$

$Z = -2$ 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.

Answer 2

(a): The required probability is $P(195 < \bar{x} < 205)=0.6826$

(b): The required probability is $P(190 < \bar{x} < 200)=0.9544$

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

$n=100$

$\mu_{\bar{x}}=200$

$\sigma{\bar{x}} =\frac{\sigma}{\sqrt{n} } \\=\frac{50}{\sqrt{100} }\\ =5$

Part(a):

within $5=200\pm 5=195,205$

$P(195 < \bar{x} < 205)=P(-1 < z < 1)\\=P(z < 1)-P(z < -1)\\=0.8413-0.1587\\=0.6826$

Part(b):

within $10=200\pm 10=190,200$

$P(190 < \bar{x} < 200)=P(-1 .98 < z < 1.98)\\=P(z < 2)-P(z < -2)\\=0.9772-0.0228\\=0.9544$

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