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Trimmed mean involves trimming $P$ percent observations from both ends.

E.g.: If you are asked to compute a 10% trimmed mean, $P = 10$.

Given a bunch of observations, $X_i$:

1. First find $n$ = number of observations.
2. Reorder them as "order statistics" $X_i$ from the smallest to the largest.
3. Find lower case $p = P/100$ = proportion trimmed.
4. Compute $n p$.

If $n p$ is an integer use $k = n p$ and trim $k$ observations at both ends.

$R$ = remaining observations = $n - 2k$.

Trimmed mean = $(1/R) \left( X_{k+1} + X_{k+2} + \ldots + X_{n-k} \right).$

Example: Find 10% trimmed mean of

2, 4, 6, 7, 11, 21, 81, 90, 105, 121

Here, $n = 10, p = 0.10, k = n p = 1$ which is an integer so trim exactly one observation at each end, since $k = 1$. Thus trim off 2 and 121. We are left with $R = n - 2k = 10 - 2 = 8$ observations.

10% trimmed mean= (1/8) * (4 + 6 + 7 + 11 + 21 + 81 + 90 + 105) = 40.625

If $ n p$ has a fractional part present, trimmed mean is a bit more complicated. In the above example, if we wanted 15% trimmed mean, $P = 15, p = 0.15, n = 10, k = n p = 1.5$. This has integer part 1 and fractional part 0.5 is present. $R = n - 2k = 10 - 2 * 1.5 = 10 - 3 = 7$. Thus $R = 7$ observations are retained.

Addendum upon @whuber's comment: To remain unbiased (after removing 2 and 121), it seems we must remove half of the 4 and half of the 105 for a trimmed mean of $(4/2 + 6 + 7 + 11 + 21 + 81 + 90 + 105/2)/7 = 34.64$

Source: Class notes on P percent trimmed mean

Addendum to the addendum: the actual 15% trimmed mean is $38.64$, not $34.64$

Trimmed mean involves trimming $P$ percent observations from both ends.

E.g.: If you are asked to compute a 10% trimmed mean, $P = 10$.

Given a bunch of observations, $X_i$:

1. First find $n$ = number of observations.
2. Reorder them as "order statistics" $X_i$ from the smallest to the largest.
3. Find lower case $p = P/100$ = proportion trimmed.
4. Compute $n p$.

If $n p$ is an integer use $k = n p$ and trim $k$ observations at both ends.

$R$ = remaining observations = $n - 2k$.

Trimmed mean = $(1/R) \left( X_{k+1} + X_{k+2} + \ldots + X_{n-k} \right).$

Example: Find 10% trimmed mean of

2, 4, 6, 7, 11, 21, 81, 90, 105, 121

Here, $n = 10, p = 0.10, k = n p = 1$ which is an integer so trim exactly one observation at each end, since $k = 1$. Thus trim off 2 and 121. We are left with $R = n - 2k = 10 - 2 = 8$ observations.

10% trimmed mean= (1/8) * (4 + 6 + 7 + 11 + 21 + 81 + 90 + 105) = 40.625

If $ n p$ has a fractional part present, trimmed mean is a bit more complicated. In the above example, if we wanted 15% trimmed mean, $P = 15, p = 0.15, n = 10, k = n p = 1.5$. This has integer part 1 and fractional part 0.5 is present. $R = n - 2k = 10 - 2 * 1.5 = 10 - 3 = 7$. Thus $R = 7$ observations are retained.

Addendum upon @whuber's comment: To remain unbiased (after removing 2 and 121), it seems we must remove half of the 4 and half of the 105 for a trimmed mean of $(4/2 + 6 + 7 + 11 + 21 + 81 + 90 + 105/2)/7 = 38.64$

Source: Class notes on P percent trimmed mean

Trimmed mean involves trimming $P$ percent observations from both ends.

E.g.: If you are asked to compute a 10% trimmed mean, $P = 10$.

Given a bunch of observations, $X_i$:

1. First find $n$ = number of observations.
2. Reorder them as "order statistics" $X_i$ from the smallest to the largest.
3. Find lower case $p = P/100$ = proportion trimmed.
4. Compute $n p$.

If $n p$ is an integer use $k = n p$ and trim $k$ observations at both ends.

$R$ = remaining observations = $n - 2k$.

Trimmed mean = $(1/R) \left( X_{k+1} + X_{k+2} + \ldots + X_{n-k} \right).$

Example: Find 10% trimmed mean of

2, 4, 6, 7, 11, 21, 81, 90, 105, 121

Here, $n = 10, p = 0.10, k = n p = 1$ which is an integer so trim exactly one observation at each end, since $k = 1$. Thus trim off 2 and 121. We are left with $R = n - 2k = 10 - 2 = 8$ observations.

10% trimmed mean= (1/8) * (4 + 6 + 7 + 11 + 21 + 81 + 90 + 105) = 40.625

If $ n p$ has a fractional part present, trimmed mean is a bit more complicated. In the above example, if we wanted 15% trimmed mean, $P = 15, p = 0.15, n = 10, k = n p = 1.5$. This has integer part 1 and fractional part 0.5 is present. $R = n - 2k = 10 - 2 * 1.5 = 10 - 3 = 7$. Thus $R = 7$ observations are retained.

Addendum upon @whuber's comment: To remain unbiased (after removing 2 and 121), it seems we must remove half of the 4 and half of the 105 for a trimmed mean of $(4/2 + 6 + 7 + 11 + 21 + 81 + 90 + 105/2)/7 = 34.64$

Source: Class notes on P percent trimmed mean

Trimmed mean involves trimming $P$ percent observations from both ends.

E.g.: If you are asked to compute a 10% trimmed mean, $P = 10$.

Given a bunch of observations, $X_i$:

1. First find $n$ = number of observations.
2. Reorder them as "order statistics" $X_i$ from the smallest to the largest.
3. Find lower case $p = P/100$ = proportion trimmed.
4. Compute $n p$.

If $n p$ is an integer use $k = n p$ and trim $k$ observations at both ends.

$R$ = remaining observations = $n - 2k$.

Trimmed mean = $(1/R) \left( X_{k+1} + X_{k+2} + \ldots + X_{n-k} \right).$

Example: Find 10% trimmed mean of

2, 4, 6, 7, 11, 21, 81, 90, 105, 121

Here, $n = 10, p = 0.10, k = n p = 1$ which is an integer so trim exactly one observation at each end, since $k = 1$. Thus trim off 2 and 121. We are left with $R = n - 2k = 10 - 2 = 8$ observations.

10% trimmed mean= (1/8) * (4 + 6 + 7 + 11 + 21 + 81 + 90 + 105) = 40.625

If $ n p$ has a fractional part present, trimmed mean is a bit more complicated. In the above example, if we wanted 15% trimmed mean, $P = 15, p = 0.15, n = 10, k = n p = 1.5$. This has integer part 1 and fractional part 0.5 is present. $R = n - 2k = 10 - 2 * 1.5 = 10 - 3 = 7$. Thus $R = 7$ observations are retained.

Addendum upon @whuber's comment: To remain unbiased (after removing 2 and 121), it seems we must remove half of the 4 and half of the 105 for a trimmed mean of $(4/2 + 6 + 7 + 11 + 21 + 81 + 90 + 105/2)/7 = 34.64$

Source: Class notes on P percent trimmed mean

Addendum to the addendum: the actual 15% trimmed mean is $38.64$, not $34.64$

Trimmed mean involves trimming P percent observations from both ends.

e.g.: If your are asked to compute 10% trimmed mean, P = 10

Given a bunch of observations, $X_i$ (denote subscript as underscore)

1. First find n=number of observations
2. Reorder them as "order statistics" $X_i$ from the smallest to the largest.
3. Find lower case p = P/100 = proportion trimmed
4. Compute np (n times small p)

If np is an integer use k=np and trim k observations at both ends.

R = remaining observations = n - 2k

trimmed mean = (1/R) * [ $X_(k+1)$ + $X_(k+2)$ + ... + $X_(n-k)$ ]

Example 1: Find 10% trimmed mean of

2, 4, 6, 7, 11, 21, 81, 90, 105, 121
n = 10, p = 0.10, k = np = 1

which is an integer so trim exactly one observation at each end, since k = 1. Thus trim off 2 and 121 we are left with R = n - 2k = 10 - 2 = 8 observations.

10% trimmed mean= (1/8) * (4 + 6 + 7 + 11 + 21 + 81 + 90 + 105) = 40.625

If np has a fractional part present, trimmed mean is a bit more complicated. In the above example, if we wanted 15% trimmed mean, P = 15, p = 0.15, n = 10, k = np = 1.5. This has integer part 1 and fractional part 0.5 is present. R = n - 2k = 10 - 2 * 1.5 = 10 - 3 = 7 Thus R = 7 observations are retained.

Addendum upon @whuber's comment: To remain unbiased (after removing 2 and 121), it seems we must remove half of the 4 and half of the 105 for a trimmed mean of (4/2 + 6 + 7 + 11 + 21 + 81 + 90 + 105/2)/7 = 34.64

Source: Class notes on P percent trimmed mean

Trimmed mean involves trimming $P$ percent observations from both ends.

E.g.: If you are asked to compute a 10% trimmed mean, $P = 10$.

Given a bunch of observations, $X_i$:

1. First find $n$ = number of observations.
2. Reorder them as "order statistics" $X_i$ from the smallest to the largest.
3. Find lower case $p = P/100$ = proportion trimmed.
4. Compute $n p$.

If $n p$ is an integer use $k = n p$ and trim $k$ observations at both ends.

$R$ = remaining observations = $n - 2k$.

Trimmed mean = $(1/R) \left( X_{k+1} + X_{k+2} + \ldots + X_{n-k} \right).$

Example: Find 10% trimmed mean of

2, 4, 6, 7, 11, 21, 81, 90, 105, 121

Here, $n = 10, p = 0.10, k = n p = 1$ which is an integer so trim exactly one observation at each end, since $k = 1$. Thus trim off 2 and 121. We are left with $R = n - 2k = 10 - 2 = 8$ observations.

10% trimmed mean= (1/8) * (4 + 6 + 7 + 11 + 21 + 81 + 90 + 105) = 40.625

If $ n p$ has a fractional part present, trimmed mean is a bit more complicated. In the above example, if we wanted 15% trimmed mean, $P = 15, p = 0.15, n = 10, k = n p = 1.5$. This has integer part 1 and fractional part 0.5 is present. $R = n - 2k = 10 - 2 * 1.5 = 10 - 3 = 7$. Thus $R = 7$ observations are retained.

Addendum upon @whuber's comment: To remain unbiased (after removing 2 and 121), it seems we must remove half of the 4 and half of the 105 for a trimmed mean of $(4/2 + 6 + 7 + 11 + 21 + 81 + 90 + 105/2)/7 = 34.64$

Source: Class notes on P percent trimmed mean