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:
[tex]z=\frac{x-m}{\frac{s}{\sqrt{n} } }[/tex]
For x = 78.3, the z score is:
[tex]z=\frac{x-m}{\frac{s}{\sqrt{n} } }=\frac{78.3-80}{\frac{14}{\sqrt{49} } } =-0.85[/tex]
For x = 85.1, the z score is:
[tex]z=\frac{x-m}{\frac{s}{\sqrt{n} } }=\frac{85.1-80}{\frac{14}{\sqrt{49} } } =2.55[/tex]
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:
P(78.3 < x' < 85.1) = 0.7969
Step-by-step explanation:
Given:
Sample size, n = 49
mean, u = 80
Standard deviation [tex] \sigma [/tex] = 14
Sample mean, ux' = population mean = 80
Let's find the sample standard deviation using the formula:
[tex] \sigma \bar x = \frac{\sigma}{\sqrt{n}} [/tex]
[tex] = \frac{14}{\sqrt{49}} = \frac{14}{7} = 2 [/tex]
To find the probability that the sample has a sample average between 78.3 and 85.1, we have:
[tex] 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}] [/tex]
= 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
