Question

A survey indicates that for each trip to the supermarket, a shopper spends an average of 45 minutes with a standard deviation of 12 minutes in the store. The length of time spent in the store is normally distributed and is represented by the variable x. A shopper enters the store.

1. Find the Z scores for the 2 Variables 35 and 65. Z 65 =

   Z 35 =

2. Probability that the shopper will be in the store for between 35 and 65 minutes.

   P =

3. Find the probability that the shopper will be in the store less than 25 minutes.

P =

4. Find the probability that the shopper will be in the store more than 50 minutes.

P = 0.3384

5. If 200 shoppers enter the store, how many shoppers would you expect to be in the store more than 40 minutes?

   P = _________

These questions do not have anything to do with the above information

6. Find the z-score that corresponds to P₁₅.

Z = _________

7. Find the z-score that corresponds to a cumulative area of 0.4520.

Z = _________

8. Find the z-score that has 15.75% of the distribution’s area to its right.

Z = __________

9. Scores for the California Peace Officer Standards and Training test are normally distributed, with a mean of 75 and a standard deviation of 8. An agency will only hire applicants with scores in the top 5%. What is the lowest score you can earn and still be eligible to be hired by the agency?

Score = _________

A survey indicates that for each trip to the Target store, a shopper spends an average of 25 minutes with a standard deviation of 14 minutes in the store. The length of time spent in the store is normally distributed and is represented by the variable x. A shopper enters the store. Find the probability that the shopper will be in the store for between 25 and 50 minutes.

10. Probability that the shopper will be in the store for between 25 and 50 minutes.

   P = _________

11. Find the probability that the shopper will be in the store less than 25 minutes. distribution

P = _________

12. Find the probability that the shopper will be in the store more than 30 minutes. distribution.

   P = _________

13. If 200 shoppers enter the store, how many shoppers would you expect to be in the store more than 40 minutes?

Shoppers = _________

Answer 1

#1) mean ( µ ) = 45 and standard deviation (σ) = 12

Find the Z scores for the 2 Variables 35 and 65.

Z 65 == = 1.67

   Z35 = = = -0.83

Probability that the shopper will be in the store for between 35 and 65 minutes.

z score table value corresponding to z = 1.67 is 0.9525

z score table value corresponding to z = 1.67 is 0.2033

Probability = 0.9525 - 0.2033

P = 0.7493

#2) Find the probability that the shopper will be in the store less than 25 minutes.

Z == = -1.67

P = 0.0475

#3) Find the probability that the shopper will be in the store more than 50 minutes.

Z == = 0.42

z score table value corresponding to z = 0.42 is 0.6628

P = 1 - 0.6628

P = 0.3372

#4) If 200 shoppers enter the store, how many shoppers would you expect to be in the store more than 40 minutes?

  

Z == = -0.42

z score table value corresponding to z = -0.42 is 0.3372

P = 1 - 0.3372

P = 0.6628

Expected shoppers = 200*0.6628 = 133

#5) Find the z-score that corresponds to P15.

We have to find z score corresponding to area 0.1500 on the z score table

Z = -1.04 --- (from z score table )

#6) Find the z-score that corresponds to a cumulative area of 0.4520.

We have to find z score corresponding to area 0.4520 on the z score table

Z = -0.12 --- (from z score table )

#7) Find the z-score that has 15.75% of the distribution’s area to its right.

1-0.1575 = 0.8425

We have to find z score corresponding to area 0.8425 on the z score table

Z = 1.00 --- (from z score table )

#8) mean of 75 and a standard deviation of 8. An agency will only hire applicants with scores in the top 5%. What is the lowest score you can earn and still be eligible to be hired by the agency?

1 - 0.05 = 0.95

We have to find z score corresponding to area 0.9500 on the z score table

z = 1.645

x = SD*z + mean = (8*1.645) + 75

Score = 88.16