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میتونی لایو بذاری!

a) Give a reason why sampling would be required in order to reach a conclusion about

(i) the mean height of adult males in England,

(ii) the mean weight that can be supported by a single cable of a certain type without the cable breaking.

b) The weights, in kg, of sacks of potatoes are represented by the random variable $X$ with mean $\mu $ and standard deviation $\sigma $. The weights of a random sample of 500 sacks of potatoes are found and the results are summarised below.

$n = 500$,  $\sum {x = 9850} $,  $\sum {{x^2} = 194125} $.

(i) Calculate unbiased estimates of $\mu $ and ${\sigma ^2}$.

(ii) A further random sample of 60 sacks of potatoes is taken. Using your values from part (b) (i), find the probability that the mean weight of this sample exceeds 19.73 kg.

(iii) Explain whether it was necessary to use the Central Limit Theorem in your calculation in part (b) (ii).

پاسخ تشریحی :
نمایش پاسخ

a)(i) Pop too large

Not all pop accessible

(ii) Testing involves destruction

b)(i) ${\raise0.5ex\hbox{$\scriptstyle {9850}$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle {500}$}} = \left( {19.5} \right)$

${\raise0.5ex\hbox{$\scriptstyle {500}$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle {499}$}}\left( {{\raise0.5ex\hbox{$\scriptstyle {194125}$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle {500}$}} - {{\left( {{\raise0.5ex\hbox{$\scriptstyle {9850}$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle {500}$}}} \right)}^2}} \right)$

$ = 0.160\left( {32} \right){\text{ }}\left( {3{\text{ }}sfs} \right)$ or $80/499$

(ii) $\frac{{19.73 - 19.7}}{{\sqrt {\frac{{''0.160''}}{{60}}} }}$

$ = 0.580$ or $0.581$

$1 - \Phi \left( {''0.580''} \right)$

$\left( { = 1 - 0.7191} \right)$

$ = 0.281$

(iii) “Yes” must be seen or implied to gain mks $X$ not nec’y normal

Sample large

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