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Solve the following problem using Python, implementing the functions described below, one line at a time
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Description:
 def plot_qq_exp(fignum, I, title, subplot=False):
 """ plots data against an exponential distribution in 0=>90. Parameters _________ fignum : matplotlib figure number I : data title : plot title subplot : boolean, if True plot as subplot with 1 row, two columns with fignum the plot number """

    if subplot == True:
        plt.subplot(1, 2, fignum)
    else:
        plt.figure(num=fignum)
    X, Y, dpos, dneg = [], [], 0., 0.
    rad = old_div(np.pi, 180.)
    xsum = 0
    for i in I:
        theta = (90. - i) * rad
        X.append(1. - np.cos(theta))
        xsum += X[-1]
    X.sort()
    n = float(len(X))
    kappa = old_div((n - 1.), xsum)
    for i in range(len(X)):
        p = old_div((float(i) - 0.5), n)
        Y.append(-np.log(1. - p))
        f = 1. - np.exp(-kappa * X[i])
        ds = old_div(float(i), n) - f
        if dpos < ds:
            dpos = ds
        ds = f - old_div((float(i) - 1.), n)
        if dneg < ds:
            dneg = ds
    if dneg > dpos:
        ds = dneg
    else:
        ds = dpos
    Me = (ds - (old_div(0.2, n))) * (np.sqrt(n) + 0.26 +
                                     (old_div(0.5, (np.sqrt(n)))))  # Eq. 5.15 from Fisher et al. (1987)

    plt.plot(Y, X, 'ro')
    bounds = plt.axis()
    plt.axis([0, bounds[1], 0., bounds[3]])
    notestr = 'N: ' + '%i' % (n)
    plt.text(.1 * bounds[1], .9 * bounds[3], notestr)
    notestr = 'Me: ' + '%7.3f' % (Me)
    plt.text(.1 * bounds[1], .8 * bounds[3], notestr)
    if Me > 1.094:
        notestr = "Not Exponential"
    else:
        notestr = "Exponential (95%)"
    plt.text(.1 * bounds[1], .7 * bounds[3], notestr)
    plt.title(title)
    plt.xlabel('Exponential Quantile')
    plt.ylabel('Data Quantile')
    return Me, 1.094