Calculating histogram function, $h_X(75.5)$ and $int_{54.2}^{67.8} h_X(x)$
$begingroup$
A population $prod$ and a measure $X$ on $prod$ produced the data set below
$71, 80, 58, 60, 63, 64, 51, 55, 48, 67$
Let $h_X$ denote the density histogram function of $X$ based on the four intervals
$(47, 54.2], (54.2, 63.4], (63.4, 67.8]$ and $(67.8, 81]$
$(a)$ Calculate $h_X(75.5)$
$(b)$ Calculate $int_{54.2}^{67.8} h_X(x) dx$
attempt
$(a)$ I made a table
begin{array}{|c|c|c|c|}
hline
interval& (47, 54.2] & (54.2, 63.4] & (63.4, 67.8] & (67.8, 81] \ hline
frequency & 2/10& 4/10& 2/10 & 2/10\ hline
length & 7.2 & 9.2 & 4.4 & 13.2\ hline
h_X(x) & frac{0.2}{7.2} & frac{0.4}{9.2} & frac{0.2}{4.4} & frac{0.2}{13.2} \ hline
end{array}
Therefore $h_X(75.5)$ lies in $(67.8, 81]$
$h_X(75.5) = frac{0.2}{13.2}$
(b)
$int_{54.2}^{67.8} h_X(x) = F_X(67.8) - F_X(54.2) = frac{8}{10} - frac{2}{10} = frac{6}{10}$
is this correct?
statistics
$endgroup$
add a comment |
$begingroup$
A population $prod$ and a measure $X$ on $prod$ produced the data set below
$71, 80, 58, 60, 63, 64, 51, 55, 48, 67$
Let $h_X$ denote the density histogram function of $X$ based on the four intervals
$(47, 54.2], (54.2, 63.4], (63.4, 67.8]$ and $(67.8, 81]$
$(a)$ Calculate $h_X(75.5)$
$(b)$ Calculate $int_{54.2}^{67.8} h_X(x) dx$
attempt
$(a)$ I made a table
begin{array}{|c|c|c|c|}
hline
interval& (47, 54.2] & (54.2, 63.4] & (63.4, 67.8] & (67.8, 81] \ hline
frequency & 2/10& 4/10& 2/10 & 2/10\ hline
length & 7.2 & 9.2 & 4.4 & 13.2\ hline
h_X(x) & frac{0.2}{7.2} & frac{0.4}{9.2} & frac{0.2}{4.4} & frac{0.2}{13.2} \ hline
end{array}
Therefore $h_X(75.5)$ lies in $(67.8, 81]$
$h_X(75.5) = frac{0.2}{13.2}$
(b)
$int_{54.2}^{67.8} h_X(x) = F_X(67.8) - F_X(54.2) = frac{8}{10} - frac{2}{10} = frac{6}{10}$
is this correct?
statistics
$endgroup$
add a comment |
$begingroup$
A population $prod$ and a measure $X$ on $prod$ produced the data set below
$71, 80, 58, 60, 63, 64, 51, 55, 48, 67$
Let $h_X$ denote the density histogram function of $X$ based on the four intervals
$(47, 54.2], (54.2, 63.4], (63.4, 67.8]$ and $(67.8, 81]$
$(a)$ Calculate $h_X(75.5)$
$(b)$ Calculate $int_{54.2}^{67.8} h_X(x) dx$
attempt
$(a)$ I made a table
begin{array}{|c|c|c|c|}
hline
interval& (47, 54.2] & (54.2, 63.4] & (63.4, 67.8] & (67.8, 81] \ hline
frequency & 2/10& 4/10& 2/10 & 2/10\ hline
length & 7.2 & 9.2 & 4.4 & 13.2\ hline
h_X(x) & frac{0.2}{7.2} & frac{0.4}{9.2} & frac{0.2}{4.4} & frac{0.2}{13.2} \ hline
end{array}
Therefore $h_X(75.5)$ lies in $(67.8, 81]$
$h_X(75.5) = frac{0.2}{13.2}$
(b)
$int_{54.2}^{67.8} h_X(x) = F_X(67.8) - F_X(54.2) = frac{8}{10} - frac{2}{10} = frac{6}{10}$
is this correct?
statistics
$endgroup$
A population $prod$ and a measure $X$ on $prod$ produced the data set below
$71, 80, 58, 60, 63, 64, 51, 55, 48, 67$
Let $h_X$ denote the density histogram function of $X$ based on the four intervals
$(47, 54.2], (54.2, 63.4], (63.4, 67.8]$ and $(67.8, 81]$
$(a)$ Calculate $h_X(75.5)$
$(b)$ Calculate $int_{54.2}^{67.8} h_X(x) dx$
attempt
$(a)$ I made a table
begin{array}{|c|c|c|c|}
hline
interval& (47, 54.2] & (54.2, 63.4] & (63.4, 67.8] & (67.8, 81] \ hline
frequency & 2/10& 4/10& 2/10 & 2/10\ hline
length & 7.2 & 9.2 & 4.4 & 13.2\ hline
h_X(x) & frac{0.2}{7.2} & frac{0.4}{9.2} & frac{0.2}{4.4} & frac{0.2}{13.2} \ hline
end{array}
Therefore $h_X(75.5)$ lies in $(67.8, 81]$
$h_X(75.5) = frac{0.2}{13.2}$
(b)
$int_{54.2}^{67.8} h_X(x) = F_X(67.8) - F_X(54.2) = frac{8}{10} - frac{2}{10} = frac{6}{10}$
is this correct?
statistics
statistics
asked Jan 30 at 20:04
Bas basBas bas
49512
49512
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