      exceptions will be raised.

    where : array_like of bool, optional
        Elements to include in the variance. See `~numpy.ufunc.reduce` for
        details.

        .. versionadded:: 1.20.0

    Returns
    -------
    variance : ndarray, see dtype parameter above
        If ``out=None``, returns a new array containing the variance;
        otherwise, a reference to the output array is returned.

    See Also
    --------
    std, mean, nanmean, nanstd, nanvar
    :ref:`ufuncs-output-type`

    Notes
    -----
    The variance is the average of the squared deviations from the mean,
    i.e.,  ``var = mean(x)``, where ``x = abs(a - a.mean())**2``.

    The mean is typically calculated as ``x.sum() / N``, where ``N = len(x)``.
    If, however, `ddof` is specified, the divisor ``N - ddof`` is used
    instead.  In standard statistical practice, ``ddof=1`` provides an
    unbiased estimator of the variance of a hypothetical infinite population.
    ``ddof=0`` provides a maximum likelihood estimate of the variance for
    normally distributed variables.

    Note that for complex numbers, the absolute value is taken before
    squaring, so that the result is always real and nonnegative.

    For floating-point input, the variance is computed using the same
    precision the input has.  Depending on the input data, this can cause
    the results to be inaccurate, especially for `float32` (see example
    below).  Specifying a higher-accuracy accumulator using the ``dtype``
    keyword can alleviate this issue.

    Examples
    --------
    >>> a = np.array([[1, 2], [3, 4]])
    >>> np.var(a)
    1.25
    >>> np.var(a, axis=0)
    array([1.,  1.])
    >>> np.var(a, axis=1)
    array([0.25,  0.25])

    In single precision, var() can be inaccurate:

    >>> a = np.zeros((2, 512*512), dtype=np.float32)
    >>> a[0, :] = 1.0
    >>> a[1, :] = 0.1
    >>> np.var(a)
    0.20250003

    Computing the variance in float64 is more accurate:

    >>> np.var(a, dtype=np.float64)
    0.20249999932944759 # may vary
    >>> ((1-0.55)**2 + (0.1-0.55)**2)/2
    0.2025

    Specifying a where argument:

    >>> a = np.array([[14, 8, 11, 10], [7, 9, 10, 11], [10, 15, 5, 10]])
    >>> np.var(a)
    6.833333333333333 # may vary
    >>> np.var(a, where=[[True], [True], [False]])
    4.0

    rą