Standard +0.8 This question requires understanding that the median condition gives P(X ≤ 0.8816) = 0.5, using this to find the relationship between k and n, then calculating P(X < 0.8) by integration, and finally applying binomial expectation. It involves multiple connected steps (finding k from normalization, using median to constrain n, evaluating a definite integral, then applying E(Y) = np) but each individual step uses standard techniques from Further Statistics.
7 The continuous random variable \(X\) has probability density function
\(f ( x ) = \begin{cases} k x ^ { n } & 0 \leqslant x \leqslant 1 , \\ 0 & \text { otherwise, } \end{cases}\)
where \(k\) is a constant and \(n\) is a parameter whose value is positive.
It is given that the median of \(X\) is 0.8816 correct to 4 decimal places.
Ten independent observations of \(X\) are obtained.
Find the expected number of observations that are less than 0.8 .
Equate integral of f(x), upper limit 0.8816, to 0.5
Set up equation in n or k only
Correct equation [or 0.8816k = 0.5]
Correct n or k, awrt 4.5(0) or 5.5(0)
Use their f(x), upper limit 0.8, with their n
10(their p)
Answer
Marks
NB: can get 2.93 from wrong integration
0.8816 and 0.5 reversed: (M1A1M1)
M1A0A0 (M1A1)
Not just 3, but ISW if 2.93 rounded to 3
Question 7:
7 | 1
k x n d x = 1 ⇒ k = n + 1
0
0.8816
ò k x n d x = 0 .5
0
k(0.8816)n+1/(n + 1) = 0.5 and use k = n + 1
0.8816n+1 = 0.5
n = 4.5 or k = 5.5
P(X < 0.8) = 0.85.5(= 0.293)
E(number < 0.8) = 2.93 | M1
A1
M1*
depM1
A1
A1
M1
A1ft
[8] | 2.1
1.1
1.1
3.1a
1.1
3.1a
3.4
2.2a | Equate integral of f(x), upper limit 1, to 1
Correct equation for n and k
Equate integral of f(x), upper limit 0.8816, to 0.5
Set up equation in n or k only
Correct equation [or 0.8816k = 0.5]
Correct n or k, awrt 4.5(0) or 5.5(0)
Use their f(x), upper limit 0.8, with their n
10(their p)
NB: can get 2.93 from wrong integration | 0.8816 and 0.5 reversed: (M1A1M1)
M1A0A0 (M1A1)
Not just 3, but ISW if 2.93 rounded to 3
7 The continuous random variable $X$ has probability density function\\
$f ( x ) = \begin{cases} k x ^ { n } & 0 \leqslant x \leqslant 1 , \\ 0 & \text { otherwise, } \end{cases}$\\
where $k$ is a constant and $n$ is a parameter whose value is positive.
It is given that the median of $X$ is 0.8816 correct to 4 decimal places.
Ten independent observations of $X$ are obtained.
Find the expected number of observations that are less than 0.8 .
\hfill \mbox{\textit{OCR Further Statistics 2022 Q7 [8]}}