find the probability that a family of four selected at random spends between $150 and $200

Question

The mean amount of money that a family of four will spend at Kings Island, including the food and souvenirs, is $130 with a standard deviation of $12.  Assume that this distribution is normal.

 

1. Find the probability that a particular family of four selected at random spends between $150 and $200?

 

1. What is the probability that the family spends less than $140?

 

1. What is the probability that the family spends more than $170?

 

1. What is the probability that the family spends between $130 and $190? (Be able to draw a graph to illustrate your results)

 

1. Find the cost that represents the 50th percentile.

 

1. Find the cost that represents the 90th percentile.

 

1. 5% of the families spend below what value?

 

1. The top 5% of the families spend above what value?

 

1. Between what two values will the middle 50% of the families spend?

 

1. What percent of the families spend at least $120?

 

Answer 1

1. The z score for $150 is (150-130)/12=5/3=1.67. This corresponds to a probability of about 0.0475 that the family will spend $150 or more at Kings Island. The z score for $200 is (200-130)/12=5.83, but this value is a higher deviation than 3.5, the limit of my reference table, corresponding to a probability of close to zero. Therefore $150 to $200 is legitimately covered by all values above $150.
2. Probability of less than $140 is given through a z score of (140-130)/12=0.833 corresponding to 0.7975 approx.
3. More than $170 is given by z score (170-130)/12=3.33, probability=0.0004.
4. Between $130 and $190 is 0.50, because $130 is the mean and (190-130)/12=5, off the scale for z.
5. The 50th percentile is the spending level below which half (50%) of families surveyed belong, so it's the mean=$130.
6. The z score corresponding to the 90th percentile is 1.285. If X is the score then (X-130)=1.285*12, so X=145, so 90% of families spend $145 or less.
7. For the lowest 5% spending families, corresponding to a z score of 1.645, the amount is given by X in (130-X)=1.645*12, so X=130-1.645*12=$110 approx.
8. The top spending families are above 0.95, corresponding to a z score of 1.645, associated with a spend of X=130+1.645*12=$150 approximately.
9. Mean+standard deviation=130+12=between 118 and 142.
10. $120 is $10 below the mean, with a z score of -10/12=-0.833, corresponding to 0.2024 or 20.24% of families. So 100-20.24=79.76% spend more than $120. (This is complementary to the $140 in question 2 above.)