Statistical level_2 Set 2 (Q.5).ipynb

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   "source": [
    "import pandas as pd\n",
    "import numpy as np\n",
    "from scipy import stats"
   ]
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   "source": [
    "### mean profit from the two different division is mean1 + mean2"
   ]
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     "text": [
      " mean profit Rs 540 million\n"
     ]
    }
   ],
   "source": [
    "mean = (5 + 7)\n",
    "print(\" mean profit Rs\",mean*45, 'million')"
   ]
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   "source": [
    "### Variace of profit from the two differnt division is SD = SD1 + SD2\n"
   ]
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      "Standard deviation Rs 270 million\n"
     ]
    }
   ],
   "source": [
    "SD = (3^2)*(4^2)\n",
    "print(\"Standard deviation Rs\",SD*45,'million')"
   ]
  },
  {
   "cell_type": "markdown",
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   "source": [
    "## Q2.A.\tSpecify a Rupee range (centered on the mean) such that it contains 95% probability for the annual profit of the company."
   ]
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   "outputs": [],
   "source": [
    "Range = stats.norm.interval(0.95,540,270)"
   ]
  },
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      " Rs range (10.809724174185362, 1069.1902758258148) million\n"
     ]
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   ],
   "source": [
    "print(\" Rs range\",Range,'million')"
   ]
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   "source": [
    "## B"
   ]
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   "source": [
    "for calculating percentile we use formula = X=μ + Zσ\n",
    "Z value for the 5th percentile = -1.64"
   ]
  },
  {
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   "id": "4e248697",
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   "outputs": [
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     "text": [
      "the 5th percentile of profit 97.20000000000005 million\n"
     ]
    }
   ],
   "source": [
    "X = 540+(-1.64)*(270)\n",
    "print (\"the 5th percentile of profit\",X,'million')"
   ]
  },
  {
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   "metadata": {},
   "source": [
    "# C"
   ]
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   "execution_count": null,
   "id": "831aa81a",
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   "outputs": [],
   "source": [
    "probability of making loss"
   ]
  },
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   "execution_count": 14,
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   "outputs": [
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   "source": [
    "## probability of making loss in 1 division,p(X<0)\n",
    "stats.norm.cdf(0,5,3)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 16,
   "id": "aabe8546",
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   "outputs": [
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     "execution_count": 16,
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   ],
   "source": [
    "## probability of making loss in 2 division,p(X<0)\n",
    "stats.norm.cdf(0,7,4)\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "id": "ff19a6a4",
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   "outputs": [],
   "source": [
    "from above we can infer that probability of 1st division is greater than 2nd ,so the probability of making loss in 1st division more."
   ]
  }
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