Part 1

Use below table to answer the given below questions.

**Q 1. A** Use above table to find the joint probability of the people who planned to purchase and actually placed and order.

**Q 1. B **Use above table to find the joint probability of the people who planned to purchase and actually placed and order, given that people planned to purchase.

**Q 2. **An electrical manufacturing company conducts quality checks at specified periods on the products it manufactures. Historically, the failure rate for the manufacturing item is 5%. Suppose a random sample of 10 manufactured items is selected. Answer the following questions.

**A.** Probability that none of the items are defective?

**B.** Probability that exactly one of the items is defective?

**C.** Probability that two or fewer of the items are defective?

**D.** Probability that three or more of the items are defective?

**Q3. **A car salesman sells on an average 3 cars per week.

**A. **What is probability that in a given week he will sell some cars.

**B. **What is probability that in a given week he will sell 2 or more but less that 5 cars.

**C. **Plot the Poisson distribution function for cumulative probability of cars sold per-week ns number of cars sold per week.

**Q4. **Accuracy in understanding orders for a speech based bot at a restaurant is important for the Company X which has designed, marketed and launched the product for a contactless delivery due to the COVID-19 pandemic. Recognition accuracy that measures the percentage of orders that are taken correctly is 86.8%. Suppose that you place an order with the bot and two friends of your independently place orders with the same bot. Answer the following questions.

**A. **What is the probability that all three orders will be recognised correctly.

**B. **What is the probability that none of the three orders will be recognised correctly.

**C. **What is the probability that at least two of the three orders will be recognised correctly?

**Q5. **Explain 1 real life industry scenario(other than the ones mentioned above) where you can use the concepts learnt in this module of Applied statistics to get a data driven business solution.

**Code Implementation**

```
# importing libraries
import pandas as pd
import numpy as np
import warnings
import matplotlib.animation as animation
import seaborn as sns
import math
import matplotlib.pyplot as plt
sns.set(style="darkgrid")
pd.set_option('display.max_columns', 100)
pd.set_option('display.max_colwidth', -1) # data and columns are displayed correctly
pd.options.display.float_format = '{:20,.2f}'.format # display float value with correct precision
warnings.filterwarnings('ignore')
```

## Que1a

```
##You observe from the above table, that 400 people planned to purchase and actually placed an order is
##400 out of 2000 people.
plan_purchase = 400
total = 2000
joint_prob = round(plan_purchase/total,4)
joint_prob
```

Output:

0.2

##Que1b

```
# P1 = P(Actually placed an order | Planned to purchase)
# P2 = Planned to purchase and actually placed an order
# P3 = Planned to purchase
## P1 = P2/P3
P1 = (400 / 2000 )/ (500 / 2000)
```

## Que2

```
import numpy as np
import pandas as pd
import scipy.stats as stats
import matplotlib.pyplot as plt
p = 0.05 # failure rate of products that the company manufactures is 5%
n = 10 # sample size
k = np.arange(0,11) # An array of quantiles
binomial = stats.binom.pmf(k,n,p)
print(binomial)
```

Output:

[5.98736939e-01 3.15124705e-01 7.46347985e-02 1.04750594e-02 9.64808106e-04 6.09352488e-05 2.67259863e-06 8.03789063e-08 1.58642578e-09 1.85546875e-11 9.76562500e-14]

## que2a

`binomial[0]`

Output:

0.5987369392383789

## que2b

`binomial[1]`

Output:

0.31512470486230504

## que2c

```
cumbinomial = stats.binom.cdf(k,n,p)
cumbinomial[2]
```

Output:

0.9884964426207031

## que2d

```
P = 1- cumbinomial[2]
P
```

Output:

0.01150355737929687

## Que3

```
rate = 3 # which is the mean
n = np.arange(0,16)
cumpoisson = stats.poisson.cdf(n,rate)
cumpoisson
```

Output:

array([0.04978707, 0.19914827, 0.42319008, 0.64723189, 0.81526324, 0.91608206, 0.96649146, 0.9880955 , 0.99619701, 0.99889751, 0.99970766, 0.99992861, 0.99998385, 0.9999966 , 0.99999933, 0.99999988])

## que3a

```
P = 1 - cumpoisson[0]
P
```

Output:

0.950212931632136

## que3b

```
# Let P1 be the probability that the salesman sells more than 4 cars
P1 = cumpoisson[4] # P(X >=4)
# Let P2 be the probability that the salesman sells more than 1 cars
P2 = cumpoisson[1] # P(X >=1)
P = P1 - P2 # Prob. that the salesman will sell between 2 and 5 cars
P
```

Output:

0.6161149710523164

## que3c

```
poisson = stats.poisson.pmf(n,rate)
plt.plot(n,poisson, 'o-')
plt.title('Poisson')
plt.xlabel('Number of cars sold per week')
plt.ylabel('Cumulative Prob of cars sold per week')
plt.show()
```

Output:

## Que4

```
#Because there are three orders and the probability of a correct order is 0.868.
# Using Binomial distribution equation,
P(X = 3|n =3, pi given = 0.868)
3!/ 3!(3-3)! * (0.868)^3 * (1-0.868)^3-3 = 0.6540
Likewise, calculate X= 0, X=2
(X = 0) = 0.0023
(X = 2) = 0.2984
Hence, P(X>=2) = P(X=2)+P(X=3) = 0.9524
## que4a - The probability that all the three orders are recognized correctly is 0.6540, 65.4% .
## que4b - The probability that none of the orders are recognized correctly is 0.0023, 0.23% .
## que4c - The probability that atleat two of the three are recognized correctly is 0.9524, 95.24%
```

## Que5

`Insurance firms employ the theory of probability or theoretical probability when developing a policy or determining a premium rate. Probability theory is a statistical strategy for predicting the potential of future outcomes. For example, Issuing health insurance to an alcoholic is likely to be more expensive than giving health insurance to a healthy individual. Statistical study reveals that a habitual drinker faces substantial health risks, posing a significant financial risk due to the increased likelihood of severe sickness and, as a result, the need to file a claim for premium money.`

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