گاما رو نصب کن!

{{ number }}
اعلان ها
اعلان جدیدی وجود ندارد!
کاربر جدید

جستجو

پربازدیدها: #{{ tag.title }}

جستجوهای پرتکرار

میتونی لایو بذاری!
  فرم معتبر نیست.

The marks of candidates in Mathematics and English in 2009 were represented by the independent random variables $X$ and $Y$ with distributions $N\left( {28,{\text{ }}{{5.6}^2}} \right)$ and $N\left( {52,{\text{ }}{{12.4}^2}} \right)$ respectively. Each candidate’s marks were combined to give a final mark $F$, where $F = X + \frac{1}{2}Y$.

a) Find $E\left( F \right)$ and $Var\left( F \right)$.

b) The final marks of a random sample of 10 candidates from Grinford in 2009 had a mean of 49. Test at the 5% significance level whether this result suggests that the mean final mark of all candidates from Grinford in 2009 was lower than elsewhere.

پاسخ تشریحی :
نمایش پاسخ

a) $E\left( F \right) = 28 + 1/2 \times 52 = 54$

$Var\left( F \right) = {5.6^2} + 1/4 \times {12.4^2}$

$ = 69.8$

b) ${H_0}:$ Grinford mean $ = 54$;

${H_1};$ Grinford mean $ < 54$

$\frac{{49 - 54}}{{\sqrt {\frac{{69.8}}{{10}}} }}$

$ =  - 1.89\left( 3 \right)$ or $ - 1,89\left( 2 \right)$ allow +

Comp with $ - 1.645$ (or $1.893$ with $1.645$)

Evidence that Grinford mean lower

تحلیل ویدئویی تست

تحلیل ویدئویی برای این تست ثبت نشده است!