Plz help. I dont know how to solve this plz.
2 answers:
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Answer:
1) The standard deviation of the sampling distribution is:
√ V(X[bar])= 0.3033
2) 95% of the values on the normal distribution are between μ ± 0.6066
3) The formula for the confidence interval is:
[284.5; 285.5]
Step-by-step explanation:
Hello!
Your study variable is
X: "score obtained on the test by an eighth-grader"
This variable has normal distribution:
X~N(μ;σ²)
Now the since the sample mean, X[bar], is obtained from n samples of the study variable, it is also a variable and has the same distribution as the original variable with the difference that is conditioned by the number of n samples from which it was calculated. So its distribution is:
X[bar]~N(μ;σ²/n)
Where E(X[bar])=μ and V(X[bar])= σ²/n
The sample information is:
n= 170100
X[bar]= 285
And the population standard deviation is known to be σ= 125.1
1) The standard deviation of the sampling distribution is:
√ V(X[bar]) = √(σ²/n) = σ/√n = 125.1/√170100 = 0.3033
2) 95% of the values on the normal distribution are between μ ± 2σ
This is μ ± 2*0.3033 = μ ± 0.6066
3) The formula for the confidence interval is:
X[bar] ± *σ/√n
285 ± 1.64 * 0.3033
[284.5; 285.5]
I hope it helps!