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UGC-NET-Paper1

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If 'Z’ represents the standard normal variate, then arrange the following probabilities in increasing order:

A. P (0≤Z≤1)
B. P (-∞≤Z≤0)
C. P(1≤Z≤3)
D. P(2≤Z≤+w)

1. ABCD
2. BCDA
3. CDAB
4. DCAB

यदि 'z' मानक सामान्य चर का प्रतिनिधित्व करता है, तो बढ़ते क्रम में निम्नलिखित संभावनाओं की व्यवस्था करें:

A. P (0≤Z≤1)
B. P (-∞≤Z≤0)
C. P(1≤Z≤3)
D. P(2≤Z≤+w)

1. ABCD
2. BCDA
3. CDAB
4. DCAB

This Question came in

UGC-NET-Commerce-07-January-2025-Shift-1-Q23

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Detailed Explanation & Answer

Let’s estimate each probability under the standard normal distribution (mean = 0, standard deviation = 1):

D. P(2 ≤ Z ≤ ∞) ≈ very small (about 0.0228) → smallest
C. P(1 ≤ Z ≤ 3) ≈ 0.8413 − 0.3413 = 0.34
A. P(0 ≤ Z ≤ 1) ≈ 0.3413
B. P(−∞ ≤ Z ≤ 0) = 0.5

आइए मानक सामान्य वितरण (माध्य = 0, मानक विचलन = 1) के तहत प्रत्येक संभावना का अनुमान लगाएं:

D. P(2 ≤ Z ≤ ∞) ≈ बहुत छोटा (लगभग 0.0228) → सबसे छोटा
C. P(1 ≤ Z ≤ 3) ≈ 0.8413 − 0.3413 = 0.34
A. P(0 ≤ Z ≤ 1) ≈ 0.3413
B. P(−∞ ≤ Z ≤ 0) = 0.5

The Correct Answer is 4

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