The seating capacity of a stadium is 65,500. So far, 49,632

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 estimat

zmey [24]

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.

7 0

3 years ago

Y=1/2x + 3/4

aliya0001 [1]

0,3/4
1/2,1
are the points

4 0

3 years ago

8. Let X be the number of cars per minute passing a certain point of some road between 8am and 10am on a Sunday. Assume that X h

Lana71 [14]

Answer:

P (3 or fewer) =0.2650

Step-by-step explanation:

Mean = x` = 5

The Poisson distribution formula is given by

P(X) = e-ˣ` x`ˣ/ x!

The mean is 5 and the X takes the values 0,1,2,and 3 which means 3 or fewer, so we add the probability of all the values of X to get the desired Value of X.

P(3 or fewer ) = e-⁵ (5)³/3!+ e-⁵ (5)²/2! +e-⁵ (5)/1!+e-⁵ (5)⁰/0!

Putting the Values

P (3 or fewer) = 0.006737 . 125 / 6 + 0.006737 . 25 / 2 +0.006737 . 5 / 1 + 0.006737 . 1 / 1

P (3 or fewer) = 0.140374 + 0.08422+ 0.03369 +0.006737

P (3 or fewer) =0.2650

7 0

3 years ago

Which expression uses the greatest common factor and the distributive property to rewrite the sum 42+72

Komok [63]

Well since the GCF of 6, (6x7)+(6x12) would be the distributive property, which would be 114

8 0

3 years ago

Will you check my math its supposed to be a face!

LUCKY_DIMON [66]

It looks like a face to me

6 0

3 years ago