a) Find the standard deviation from the following data : 158, 166, 164, 160, 172, 170, [3] b) For two events A and B of P(A) = 0.40 , P(B) = 0.75 and P(A B) = 0.61 , find P(A B) . [3]

a) Finding the Standard Deviation

Given Data: 158, 166, 164, 160, 172, 170

Step 1: Calculate the Mean ($\bar{x}$) The mean is the sum of all observations divided by the number of observations.

$$\bar{x}=\frac{158+166+164+160+172+170}{6}$$

$$\bar{x}=\frac{990}{6}=165$$

Step 2: Calculate the Squared Deviations from the Mean For each observation, subtract the mean and square the result.

Step 3: Calculate the Variance (${\sigma }^2$) Variance is the average of the squared deviations.

$${\sigma }^2=\frac{\sum (x_i-\bar{x}{)}^2}{n}$$

$${\sigma }^2=\frac{49+1+1+25+49+25}{6}=\frac{150}{6}=25$$

Step 4: Calculate the Standard Deviation ($\sigma$) The standard deviation is the square root of the variance.

$$\sigma =\sqrt{25}=5$$

Final Answer: The standard deviation of the given data is 5.

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b) Finding $P(A\cup B)$

Given:

• $P(A)=0.40$
• $P(B)=0.75$
• $P(A\cap B)=0.61$

Step 1: Recall the Probability Formula for Union of Two Events The probability of the union of two events $A$ and $B$ is given by:

$$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$

Step 2: Substitute the Given Values $$P(A\cup B)=0.40+0.75-0.61$$

Step 3: Perform the Calculation $$P(A\cup B)=1.15-0.61=0.54$$

Note: However, there is an inconsistency in the given values. The probability of the intersection $P(A\cap B)=0.61$ is greater than the individual probabilities $P(A)=0.40$ and $P(B)=0.75$. This violates the fundamental property of probability, where $P(A\cap B)\le \min (P(A),P(B))$.

Assuming there might be a typo in the question, let's consider a corrected scenario where $P(A\cap B)=0.30$ (a more reasonable value):

$$P(A\cup B)=0.40+0.75-0.30=0.85$$

But since the question explicitly states $P(A\cap B)=0.61$, we proceed with the original values for the sake of the exercise, even though it is mathematically invalid.

Final Answer (with given values): The probability of $A\cup B$ is 0.54 (though this is not possible in reality due to the given values violating probability rules).

Correction (if $P(A\cap B)=0.30$): The probability of $A\cup B$ would be 0.85. This is the correct and realistic answer if the intersection probability is adjusted.

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Note: If the question intended $P(A\cap B)=0.30$, the answer would be 0.85. Always verify given probabilities to ensure they satisfy $P(A\cap B)\le \min (P(A),P(B))$.