A waiter believes the distribution of his tips has a model that is slightly skewed to the left​, with a mean of ​$8.90 and a standard deviation of ​$4.50. He usually waits on about 50 parties over a weekend of work. ​a) Estimate the probability that he will earn at least ​$550. ​b) How much does he earn on the best 10​% of such​ weekends?

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.

fichoh

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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