Respuesta :
Answer:
A. 0.3204 B. $14.669
Explanation:
Mean = 8.9 SD = 4.5
Required probability = P (X >/= 550/50)
P(X>/=11) = 1 - P[(X - mean/SD) < (11 - mean)/SD]
= 1 - P(Z < (11-8.9)/4.5)
P(X>/=11) = 1 - P(Z < 0.4666667)
Using Excel NORMDIST(0.4666667,0,1,1)
P(X>/=11) = 1 - 0.6796 = 0.3204
The probability that she will earn at least $550 = 0.3204
b. P ( X > x ) = 0.10
1 − P ( X − mean)/SD ≤ (x − mean) /SD = 0.10
P ( Z ≤ z ) = 0.90
Where,
z = (x − mean )/SD
Excel function for the value of z:
=NORMSINV(0.9)
=1.282
Hence (x - mean)/SD = 1.282
= (x - 8.9)/4.5 = 1.282
x = (1.282*4.5) + 8.9
x = 14.669
He earns $14.669 on the best 10% of such weekends.
Using the central limit theorem, the eatimated probability of earning atleast $550 and the amount earned on the best 10% of weekends is 0.00048 and $485.793 respectively.
Using the central limit theorem since sample size, n > 30 ;
- Sampling distribution of tips using the normal distribution :
σ = √(sample size) × standard deviation= √(50) ×4.50 = 31.82
μ = (sample size × mean) = 50 × 8.90 = 445
Using the Zscore relation :
- P(Z < z) = Z < (x - μ) /σ))
P(Z ≥ z) = P(550 - 445) /31.82))
P(Z ≥ z) = 3.30
Using a normal distribution table :
P(Z ≥ 3.30) = 1 - P(Z ≤ 3.30)
P(Z ≥ 3.30) = 1 - 0.99952
P(Z ≥ 3.30) = 0.00048
2.) Probability of earning on the best 10%:
- P(Z < z) = 1 - 0.10 = 0.90
P(Z < z) = 0.90
Using the normal distribution table :
Zscore at 0.90 = 1.282
Zscore = (x - μ) / σ
1.282 = (x - 445) / 31.82
31.82 × 1.282 = x - 445
x - 445 = 40.793
x = 40.793 + 445
x = $485.793
Hence, amount earned on the best 10% of such weekends is $485.793.
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