APL(NARS), 26 chars

{m←÷≢⍵⋄d÷√m×+/×⍨d←⍵-m×+/⍵}    

test:

  f←{m←÷≢⍵⋄d÷√m×+/×⍨d←⍵-m×+/⍵}
  f 1 2
¯1 1 
  f 1 2 3
¯1.224744871 0 1.224744871 
  f ¯3 1 4 1 5
¯1.642857143 ¯0.2142857143 0.8571428571 ¯0.2142857143 1.214285714 

Tcl, 115 bytes

proc S L {lmap c $L {expr ($c-[set m ([join $L +])/[set n [llength $L]].])/sqrt((([join $L -$m)**2+(]-$m)**2)/$n)}}

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CASIO BASIC (CASIO fx-9750GIII), 20 bytes

?→List1
1-Variable List1
(List1-x̄)÷σx

builtins

Factor, 34 bytes

[ dup demean swap 0 std-ddof v/n ]

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Sadly, while Factor has the z-score word, it uses the sample standard deviation instead of the population standard deviation.

Charcoal, 25 19 bytes

≧⁻∕ΣθＬθθＩ∕θ₂∕ΣＸθ²Ｌθ

Try it online! Link is to verbose version of code. Explanation:

       θ    Input array
≧           Update each element
 ⁻          Subtract
   Σ        Sum of
    θ       Input array
  ∕         Divided by
     Ｌ      Length of
      θ     Input array

Calculate \$\mu\$ and vectorised subtract it from each \$x_i\$.

  θ         Updated array
 ∕          Vectorised divided by
   ₂        Square root of
     Σ      Sum of
       θ    Updated array
      Ｘ     Vectorised to power
        ²   Literal 2
    ∕       Divided by
         Ｌ  Length of
          θ Array
Ｉ           Cast to string
            Implicitly print each element on its own line.

Calculate \$\sigma\$, vectorised divide each \$x_i\$ by it, and output the result.

Edit: Saved 6 bytes thanks to @ASCII-only for a) using SquareRoot() instead of Power(0.5) b) fixing vectorised Divide() (it was doing IntDivide() instead) c) making Power() vectorise.

Julia 0.7, 37 bytes

a->(a-mean(a))/std(a,corrected=false)

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SNOBOL4 (CSNOBOL4), 229 bytes

	DEFINE('Z(A)')
Z	X =X + 1
	M =M + A<X>	:S(Z)
	N =X - 1.
	M =M / N
D	X =GT(X) X - 1	:F(S)
	A<X> =A<X> - M	:(D)
S	X =LT(X,N) X + 1	:F(Y)
	S =S + A<X> ^ 2 / N	:(S)
Y	S =S ^ 0.5
N	A<X> =A<X> / S
	X =GT(X) X - 1	:S(N)
	Z =A	:(RETURN)

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Link is to a functional version of the code which constructs an array from STDIN given its length and then its elements, then runs the function Z on that, and finally prints out the values.

Defines a function Z which returns an array.

The 1. on line 4 is necessary to do the floating point arithmetic properly.

MathGolf, 7 bytes

▓-_²▓√/

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Explanation

This is literally a byte-for-byte recreation of Kevin Cruijssen's 05AB1E answer, but I save some bytes from MathGolf having 1-byters for everything needed for this challenge. Also the answer looks quite good in my opinion!

▓         get average of list
 -        pop a, b : push(a-b)
  _       duplicate TOS
   ²      pop a : push(a*a)
    ▓     get average of list
     √    pop a : push(sqrt(a)), split string to list
      /   pop a, b : push(a/b), split strings

Pyth, 21 19 bytes

mc-dJ.OQ@.Om^-Jk2Q2

Try it online here.

mc-dJ.OQ@.Om^-Jk2Q2Q   Implicit: Q=eval(input())
                       Trailing Q inferred
    J.OQ               Take the average of Q, store the result in J
           m     Q     Map the elements of Q, as k, using:
             -Jk         Difference between J and k
            ^   2        Square it
         .O            Find the average of the result of the map
        @         2    Square root it
                       - this is the standard deviation of Q
m                  Q   Map elements of Q, as d, using:
  -dJ                    d - J
 c                       Float division by the standard deviation
                       Implicit print result of map

Edit: after seeing Kevin's answer, changed to use the average builtin for the inner results. Previous answer: mc-dJ.OQ@csm^-Jk2QlQ2

05AB1E, 9 bytes

ÅA-DnÅAt/

Port of @Arnauld's JavaScript answer, so make sure to upvote him!

Try it online or verify all test cases.

Explanation:

ÅA          # Calculate the mean of the (implicit) input
            #  i.e. [-3,1,4,1,5] → 1.6
  -         # Subtract it from each value in the (implicit) input
            #  i.e. [-3,1,4,1,5] and 1.6 → [-4.6,-0.6,2.4,-0.6,3.4]
   D        # Duplicate that list
    n       # Take the square of each
            #  i.e. [-4.6,-0.6,2.4,-0.6,3.4] → [21.16,0.36,5.76,0.36,11.56]
     ÅA     # Pop and calculate the mean of that list
            #  i.e. [21.16,0.36,5.76,0.36,11.56] → 7.84
       t    # Take the square-root of that
            #  i.e. 7.84 → 2.8
        /   # And divide each value in the duplicated list with it (and output implicitly)
            #  i.e. [-4.6,-0.6,2.4,-0.6,3.4] and 2.8 → [-1.6428571428571428,
            #   -0.21428571428571433,0.8571428571428572,-0.21428571428571433,1.2142857142857144]

TI-Basic (83 series), 14 11 bytes

Ans-mean(Ans
Ans/√(mean(Ans²

Takes input in Ans. For example, if you type the above into prgmSTANDARD, then {1,2,3}:prgmSTANDARD will return {-1.224744871,0.0,1.224744871}.

Previously, I tried using the 1-Var Stats command, which stores the population standard deviation in σx, but it's less trouble to compute it manually.

Haskell, 80 75 68 bytes

t x=k(/sqrt(f$sum$k(^2)))where k g=g.(-f(sum x)+)<$>x;f=(/sum(1<$x))

Thanks to @flawr for the suggestions to use sum(1<$x) instead of sum[1|_<-x] and to inline the mean, @xnor for inlining the standard deviation and other reductions.

Expanded:

-- Standardize a list of values of any floating-point type.
standardize :: Floating a => [a] -> [a]
standardize input = eachLessMean (/ sqrt (overLength (sum (eachLessMean (^2)))))
  where

    -- Map a function over each element of the input, less the mean.
    eachLessMean f = map (f . subtract (overLength (sum input))) input

    -- Divide a value by the length of the input.
    overLength n = n / sum (map (const 1) input)

APL (Dyalog Unicode), 33 29 bytes

{d÷.5*⍨l÷⍨+/×⍨d←⍵-(+/⍵)÷l←≢⍵}

-4 bytes thanks to @ngn

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MATLAB, 26 bytes

Trivial-ish, std(,1) for using population standard deviation

f=@(x)(x-mean(x))/std(x,1)

APL+WIN, 41,32 30 bytes

9 bytes saved thanks to Erik + 2 more thanks to ngn

x←v-(+/v)÷⍴v←⎕⋄x÷(+/x×x÷⍴v)*.5

Prompts for vector of numbers and calculates mean standard deviation and standardised elements of input vector

APL (Dyalog Classic), 21 20 19 bytes

(-÷.5*⍨⊢÷⌹×≢)+/-⊢×≢

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⊢÷⌹ is sum of squares

⊢÷⌹×≢ is sum of squares divided by length

K (oK), 33 23 bytes

-10 bytes thanks to ngn!

{t%%(+/t*t:x-/x%#x)%#x}

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First attempt at coding (I don't dare to name it "golfing") in K. I'm pretty sure it can be done much better (too many variable names here...)

Haskell, 59 bytes

(%)i=sum.map(^i)
f l=[(0%l*y-1%l)/sqrt(2%l*0%l-1%l^2)|y<-l]

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Doesn't use libraries.

The helper function % computes the sum of ith powers of a list, which lets us get three useful values.

• 0%l is the length of l (call this n)
• 1%l is the sum of l (call this s)
• 2%l is the sum of squares of l (call this m)

We can express the z-score of an element y as

(n*y-s)/sqrt(n*v-s^2)

(This is the expression (y-s/n)/sqrt(v/n-(s/n)^2) simplified a bit by multiplying the top and bottom by n.)

We can insert the expressions 0%l, 1%l, 2%l without parens because the % we define has higher precedence than the arithmetic operators.

(%)i=sum.map(^i) is the same length as i%l=sum.map(^i)l. Making it more point-free doesn't help. Defining it like g i=... loses bytes when we call it. Although % works for any list but we only call it with the problem input list, there's no byte loss in calling it with argument l every time because a two-argument call i%l is no longer than a one-argument one g i.

Python 3 + numpy, 46 bytes

lambda a:(a-mean(a))/std(a)
from numpy import*

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J, 22 bytes

-1 byte thanks to Cows quack!

(-%[:%:1#.-*-%#@[)+/%#

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J, 31 23 bytes

(-%[:%:#@[%~1#.-*-)+/%#

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                   +/%# - mean (sum (+/) divided (%) by the number of samples (#)) 
(                 )     - the list is a left argument here (we have a hook)
                 -      - the difference between each sample and the mean
                *       - multiplied by 
               -        - the difference between each sample and the mean
            1#.         - sum by base-1 conversion
          %~            - divided by
       #@[              - the length of the samples list
     %:                 - square root
   [:                   - convert to a fork (function composition) 
 -                      - subtract the mean from each sample
  %                     - and divide it by sigma

Mathematica, 25 bytes

Mean[(a=#-Mean@#)a]^-.5a&

Pure function. Takes a list of numbers as input and returns a list of machine-precision numbers as output. Note that the built-in Standardize function uses the sample variance by default.

JavaScript (ES7),  80  79 bytes

a=>a.map(x=>(x-g(a))/g(a.map(x=>(x-m)**2))**.5,g=a=>m=eval(a.join`+`)/a.length)

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Commented

a =>                      // given the input array a[]
  a.map(x =>              // for each value x in a[]:
    (x - g(a)) /          //   compute (x - mean(a)) divided by
    g(                    //   the standard deviation:
      a.map(x =>          //     for each value x in a[]:
        (x - m) ** 2      //       compute (x - mean(a))²
      )                   //     compute the mean of this array
    ) ** .5,              //   and take the square root
    g = a =>              //   g = helper function taking an array a[],
      m = eval(a.join`+`) //     computing the mean
          / a.length      //     and storing the result in m
  )                       // end of outer map()

R, 51 45 38 37 bytes

Thanks to Giuseppe and J.Doe!

function(x)scale(x)/(1-1/sum(x|1))^.5

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Jelly, 10 bytes

_ÆmµL½÷ÆḊ×

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It's not any shorter, but Jelly's determinant function ÆḊ also calculates vector norm.

_Æm             x - mean(x)
   µ            then:
    L½          Square root of the Length
      ÷ÆḊ       divided by the norm
         ×      Multiply by that value

R + pryr, 53 52 bytes

-1 byte using sum(x|1) instead of length(x) as seen in @Robert S.'s solution

pryr::f((x-(y<-mean(x)))/(sum((x-y)^2)/sum(x|1))^.5)

For being a language built for statisticians, I'm amazed that this doesn't have a built-in function. At least not one that I could find. Even the function mosaic::zscore doesn't yield the expected results. This is likely due to using the population standard deviation instead of sample standard deviation.

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MATL, 10 bytes

tYm-t&1Zs/

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Explanation

t       % Implicit input
        % Duplicate
Ym      % Mean
-       % Subtract, element-wise
t       % Duplicate
&1Zs    % Standard deviation using normalization by n
/       % Divide, element-wise
        % Implicit display

CW for all trivial entries

Python 3 + scipy, 31 bytes

from scipy.stats import*
zscore

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Octave / MATLAB, 15 bytes

@(x)zscore(x,1)

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Jelly, 10 bytes

_Æm÷²Æm½Ɗ$

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