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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 _select_ps(p):
 # There are more generic ways of doing this but profiling # revealed that selecting these points is one of the slow # things that is easy to change. This is about 11 times # faster than the generic algorithm it is replacing. # # it is possible that different break points could yield # better estimates, but the function this is refactoring # just used linear distance. """returns the points to use for interpolating p"""

    if p >= .99:
        return .990, .995, .999
    elif p >= .975:
        return .975, .990, .995
    elif p >= .95:
        return .950, .975, .990
    elif p >= .9125:
        return .900, .950, .975
    elif p >= .875:
        return .850, .900, .950
    elif p >= .825:
        return .800, .850, .900
    elif p >= .7625:
        return .750, .800, .850
    elif p >= .675:
        return .675, .750, .800
    elif p >= .500:
        return .500, .675, .750
    else:
        return .100, .500, .675