Question

Sale amounts during lunch hour at a local subway are normally ditributed, with a mean $ 7.86, and a standard deviation og $ 1.03.

Round all of your final answers to four decimal places.

a. Find the probability that a randomly selected sale was at least $ 6.92?

b. A particular sale was $ 10.57. What is the percentile rank for this sale amount?

c. Give the sale amount that is the cutoff for the highest 85 %

d.What is the probability that a randomly selected sale is between $6.00 and $8.00?

e. What sale amount represents the cutoff for the middle 39 percent of sales?Correct ( The smaller number here)Correct(Bigger number here)

Answer 1

Solution :

Given that ,

mean = = 7.86

standard deviation = = 1.03

a. P(x 6.92)

= 1 - P(x   6.92)

= 1 - P[(x - ) / (6.92 - 7.86) / 1.03]

= 1 -  P(z -0.91)   

  Using z table,

= 1 - 0.1814

= 0.8186

Probability = 0.8186

b. P(x < 10.57)

= P[(x - ) / < (10.57 - 7.86) / 1.03]

= P(z < 2.63)

Using z table,

= 0.9957

= 99.57%

Percentile = 99.57%

c. The z-distribution of the 85% is,

P(Z > z) = 85%

= 1 - P(Z < z ) = 0.85

= P(Z < z ) = 1 - 0.85

= P(Z < z ) = 0.15

= P(Z < -1.036 ) = 0.15

z = -1.036

Using z-score formula,

x = z * +

x = -1.036 * 1.03 + 7.86

x = 6.79

Cutoff = 6.79

d. P(6.00 < x < 8.00)

= P[(6.00 - 7.86) / 1.03) < (x - ) /  < (8.00 - 7.86) / 1.03) ]

= P(-1.81 < z < 0.14)

= P(z < 0.14) - P(z < -1.81)

Using z table,

= 0.5557 - 0.0351

= 0.5206

Probability = 0.5206

e. The middle of 39%

= 1 - 39%  

= 1 - 0.39 = 0.61

/2 = 0.305

Z_/2 = Z_0.305 = -0.510

1 - /2  = 1 - 0.305 = 0.695

Z_{1 - /2} = Z_0.695  = 0.510

Using z-score formula,

x = Z_/2 * +

x = -0.510 * 1.03 + 7.86

x = 7.33

Using z-score formula,

x = Z_{1 - /2} * +

x = 0.510 * 1.03 + 7.86

x = 8.39

The cutoff for 39% middle is between 7.33 and 8.39