Algebra

Probability

Easy

applied

If X is a normal variate with mean 16 and standard deviation 4, then the value of standard normal variate Z corresponding to X = 17 is:

1.25

-1.25

-0.25

0.25

✅ Correct Option: 4

Given that $X$ is a normal variate with mean $\mu = 16$ and standard deviation $\sigma = 4$ , and we need to find the standard normal variate $Z$ corresponding to $X = 17$ .

The transformation from a normal variate to a standard normal variate uses the formula:

$Z = \dfrac{X - \mu}{\sigma}$

Substituting the given values:

$Z = \dfrac{17 - 16}{4}$

$Z = \dfrac{1}{4}$

$Z = 0.25$

This means $X = 17$ is $0.25$ standard deviations above the mean.

Therefore, the value of the standard normal variate $Z$ is $0.25$ .

2025: 14 May Shift 1

If A and B are any two events such that P(B) = P(A and B), then which of the following is correct

2022: 4 Aug Shift 1

The probability distribution of a discrete random variable X is given as :

x	0	1	2	3	4
P(X=x)	0.1	k	2k	2k	k

The value of K is:

2025: 15 May Shift 2

If a random variable $X$ has the following probability distribution:

X	0	1	2	3	4
P(X)	k	2k	3k	k²	6k²

, then

Match List-I with List-II

List-I	List-II
(A) k	(I) 3/7
(B) $P(X < 2)$	(II) 6/49
(C) $P(X > 3)$	(III) 1/7
(D) $P(2 \leq X \leq 3)$	(IV) 22/49

Choose the correct answer from the options given below:

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If $P(A) = \frac{3}{10}$ , $P(B) = \frac{2}{5}$ and $P(A \cup B) = \frac{3}{5}$ then the value of $P(B|A) + P(A|B)$ is:

2026: 29 May Shift 1

Bag A contains 3 red and 5 white balls and bag B contains 4 red and 4 white balls. A bag is selected at random, and a ball is drawn. Then the probability of getting a red ball is:

2025: 21 May Shift 2

The probability that it will rain on any particular day is 50%. The probability that it rains only on the first 4 days of the week is:

If X is a normal variate with mean 16 and standard deviation 4, then the value of standard normal variate Z corresponding to X = 17 is:

Solution

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