Question

N

2) A sample of size n= 49 is obtained. The population mean

is m= 80 and the population standard deviation is s = 14.

Find the probability that the sample has a sample average

between 78.3 and 85.1, (5 points)

-

Answer 1

Answer:

0.7969

Step-by-step explanation:

Given that: A sample of size n= 49 is obtained. The population mean  is m= 80 and the population standard deviation is s = 14.

The z score measures the number of standard deviation by which the raw sore is above or below the mean. It is given by the equation:

$z=\frac{x-m}{\frac{s}{\sqrt{n} } }$

For x = 78.3, the z score is:

$z=\frac{x-m}{\frac{s}{\sqrt{n} } }=\frac{78.3-80}{\frac{14}{\sqrt{49} } } =-0.85$

For x = 85.1, the z score is:

$z=\frac{x-m}{\frac{s}{\sqrt{n} } }=\frac{85.1-80}{\frac{14}{\sqrt{49} } } =2.55$

P(78.3<x<85.1) = P(-0.85<z<2.55) = P(z<2.55) - P(z<-0.85) = 0.9946 - 0.1977 = 0.7969

Answer 2

Answer:

P(78.3 < x' < 85.1) = 0.7969

Step-by-step explanation:

Given:

Sample size, n = 49

mean, u = 80

Standard deviation $\sigma$ = 14

Sample mean, ux' = population mean = 80

Let's find the sample standard deviation using the formula:

$\sigma \bar x = \frac{\sigma}{\sqrt{n}}$

$= \frac{14}{\sqrt{49}} = \frac{14}{7} = 2$

To find the probability that the sample has a sample average between 78.3 and 85.1, we have:

$P(78.3 < \bar x < 85.1) = \frac{P[(78.3 -80)}{2} < \frac{(\bar x - u \bar x)}{\sigma \bar x} < \frac{(85.1 -80)}{2}]$

= P( -0.85 < Z < 2.55 )

= P(Z < 2.55) - P(Z <-0.85 )

Using the standard normal table, we have:

= 0.9946 - 0.1977 = 0.7969

Approximately 0.80

Therefore, the probability that the sample has a sample average between 78.3 and 85.1 is 0.7969