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In an intelligence test administered to 1000 students, the average was 42 and standard deviation was 24. Find the number of students (i) exceeding the score 50 and (ii) between 30 and 54.

solution:
1. 𝙽𝚞𝚖𝚋𝚎𝚛 𝚘𝚏 𝚜𝚝𝚞𝚍𝚎𝚗𝚝𝚜 𝚎𝚡𝚌𝚎𝚎𝚍𝚒𝚗𝚐 𝚊 𝚜𝚌𝚘𝚛𝚎 𝚘𝚏 50
Step 1: 𝙲𝚊𝚕𝚌𝚞𝚕𝚊𝚝𝚎 𝚝𝚑𝚎 𝚉-𝚜𝚌𝚘𝚛𝚎 𝚏𝚘𝚛 𝑋 = 50
Given the mean 𝜇=42 and standard deviation 𝜎=24, the Z-score for a score 𝑋=50 is calculated as:
𝑍=\frac{𝑋−𝜇}{𝜎}=\frac{50−42}{24}=\frac{8}{24}=\frac{1}{3}≈0.33
Step 2: 𝙵𝚒𝚗𝚍 𝚝𝚑𝚎 𝚙𝚛𝚘𝚋𝚊𝚋𝚒𝚕𝚒𝚝𝚢 𝚘𝚏 𝚎𝚡𝚌𝚎𝚎𝚍𝚒𝚗𝚐 𝚊 𝚜𝚌𝚘𝚛𝚎 𝚘𝚏 50
The probability 𝑃(𝑍≥0.33) is the area to the right of 𝑍=0.33 under the standard normal curve. This can be found by subtracting the area between 𝑍=0 and 𝑍=0.33 from 0.5.
From the standard normal table, this area is approximately 0.1293.
So, 𝑃(𝑍≥0.33)=0.5−0.1293=0.3707
Step 3: 𝙲𝚊𝚕𝚌𝚞𝚕𝚊𝚝𝚎 𝚝𝚑𝚎 𝚗𝚞𝚖𝚋𝚎𝚛 𝚘𝚏 𝚜𝚝𝚞𝚍𝚎𝚗𝚝𝚜
With a total of 𝑁=1000 students, the number exceeding 50 is:
𝑁×𝑃(𝑍≥0.33)=1000×0.3707=371

2. 𝙽𝚞𝚖𝚋𝚎𝚛 𝚘𝚏 𝚜𝚝𝚞𝚍𝚎𝚗𝚝𝚜 𝚜𝚌𝚘𝚛𝚒𝚗𝚐 𝚋𝚎𝚝𝚠𝚎𝚎𝚗 30 𝚊𝚗𝚍 54
Step 1: 𝙲𝚊𝚕𝚌𝚞𝚕𝚊𝚝𝚎 𝚉-𝚜𝚌𝚘𝚛𝚎𝚜
For 𝑋=30: 𝑍_1=\frac{30−42}{24}=−0.5
For 𝑋=54: 𝑍_2=\frac{54−42}{24}=0.5
Step 2: 𝙵𝚒𝚗𝚍 𝚙𝚛𝚘𝚋𝚊𝚋𝚒𝚕𝚒𝚝𝚢 𝑃(−0.5≤𝑍≤0.5)= 0.1915.
So, 𝑃(−0.5≤𝑍≤0.5)=2×0.1915=0.383
Step 3: 𝙲𝚊𝚕𝚌𝚞𝚕𝚊𝚝𝚎 𝚗𝚞𝚖𝚋𝚎𝚛 𝚘𝚏 𝚜𝚝𝚞𝚍𝚎𝚗𝚝𝚜 𝑁×𝑃(−0.5≤𝑍≤0.5)=1000×0.383=383