- 2071 Shrawan · 6 marks
Sample mean income of 10 retailers
Short answer
Since the population is normal, sample mean is normal with mean Rs 21,000 and SE=9487/√10≈Rs 3,000. z1=-1.00 and z2=2.00, so P(18,000<x̄<27,000)=0.3413+0.4772=0.8186.
The monthly income of a particular group of retailer's follows a normal distribution with mean Rs.21,000.00 and standard deviation of Rs.9,487.00. A random sample of size 10 retailers was taken and the mean income is calculated. Find the probability that this sample lies between Rs.18,000.00 and Rs.27,000.00.
Income X ~ N(μ = Rs 21,000, σ = Rs 9,487), n = 10. Since the population is normal, x̄ is normal.
Standard error = σ/√n = 9487/√10 = 9487/3.1623 = Rs 3000 (about Rs 3,000)
- z₁ = (18,000 − 21,000)/3,000 = -1.00
- z₂ = (27,000 − 21,000)/3,000 = 2.00
P(18,000 < x̄ < 27,000) = P(−1 < Z < 2) = P(0 < Z < 1) + P(0 < Z < 2)
= 0.3413 + 0.4772 = 0.8186
Answer: the probability is about 0.8186 (81.86%).
✓ Answer checked by Claude Sonnet 5.5
Related questions
- Multiple correlation: expenditure, income, family size · Probability and Statistics, chapter 5
- Sample survey advantages; sample size for bills · Probability and Statistics, chapter 6
- Sampling distribution of mean; standard error · Probability and Statistics, chapter 4 · asked 2×
- Standard normal; normal approx of Poisson · Probability and Statistics, chapter 3 · asked 3×
- Standard normal: definition, properties, importance · Probability and Statistics, chapter 3 · asked 2×
- Normal characteristics; relation to standard normal · Probability and Statistics, chapter 3
- EPS and book value; Rs 10,00,000, 3500 shares · Project Engineering, chapter 6
- EVA; BCWS 9,50,000, BCWP 8,00,000, ACWP 10,00,000 · Project Engineering, chapter 4
- EVA application; 35 days, Rs 8,00,000 spent · Project Engineering, chapter 4
- EVA and S-curve; Rs 30,00,000, 24 months · Project Engineering, chapter 4