For this case we have that by definition, the distance between two points is given by:
We have the following points:
Substituting we have:
Thus, the distance between the points is
ANswer:
I have to take care of my sister's cat while she is on vacation. The cat is a very picky eater and will only eat a certain cat f
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Answer:
a. 0.691
b. 0.382
c. 0.933
d. $88.490
e. $58.168
f. 5th percentile: $42.103
95th percentile: $107.897
Step-by-step explanation:
We have, for the purchase amounts by customers, a normal distribution with mean $75 and standard deviation of $20.
a. This can be calculated using the z-score:
The probability that a randomly selected customer spends less than $85 at this store is 0.691.
b. We have to calculate the z-scores for both values:
The probability that a randomly selected customer spends between $65 and $85 at this store is 0.382.
c. We recalculate the z-score for X=45.
The probability that a randomly selected customer spends more than $45 at this store is 0.933.
d. In this case, first we have to calculate the z-score that satisfies P(z<z*)=0.75, and then calculate the X* that corresponds to that z-score z*.
Looking in a standard normal distribution table, we have that:
Then, we can calculate X as:
75% of the customers will not spend more than $88.49.
e. In this case, first we have to calculate the z-score that satisfies P(z>z*)=0.8, and then calculate the X* that corresponds to that z-score z*.
Looking in a standard normal distribution table, we have that:
Then, we can calculate X as:
80% of the customers will spend more than $58.17.
f. We have to calculate the two points that are equidistant from the mean such that 90% of all customer purchases are between these values.
In terms of the z-score, we can express this as:
The value for z* is ±1.64485.
We can now calculate the values for X as:
5th percentile: $42.103
95th percentile: $107.897