An srs of 380 high school seniors gained an average of ¯ x = 20.99 points in their second attempt at the sat mathematics exam. assume that the change in score has a normal distribution with standard deviation σ = 47.53 . we want to estimate the mean change in score μ in the population of all high school seniors.
(a) using the 68 – 95 – 99.7 rule or the z - table (table a), give a 95 % confidence interval ( a , b ) for μ based on this sample. (enter your answers rounded to three decimal places. if you are using crunchit, adjust the default precision under preferences as necessary. see the instructional video on how to adjust precision settings.)

Step-by-step explanation:

Hello!

Your study variable is X: scores obtained on mathematics SAT at the 2nd attempt of a high schooler.

This variable has a normal distribution and the population standard deviation is known σ= 47.53

Since you want to estimate the population mean, you have to work using the sample mean, its distribution is X[bar]~N(μ;σ²/n)

a.

Using the Z-table

[X[bar] ± *(σ/√n)]

Where:

Z_{1-\alpha /2} = Z_{0.975} = 1.96

X[bar]= 20.99

n=380

[20.99 ±1.96*(47.53/√380)]

[16.2110;25.7689]

Using the 68-95-99.7 rule, since is based on the sample, the sample mean is x[bar] and the standard deviation for the sample mean distribution is: (σ/√n)

95% is between μ ± 2σ

Under the sample mean distribution:

mean: x[bar]= 20.99

standerd deviation: σ/√n = 47.53/√380

95% is between 20.99 ± 2*(47.53/√380) = 20.99 ± 2*2.4382

[16.1136;25.8664]

I hope it helps!

yes the data was not prsented

step-by-step explanation:

the answer is 21%.