Question

Suppose it is known that the distribution of purchase amounts by customers entering a popular retail store is approximately normal with mean $75 and standard deviation $20. Round your answers to three decimal places, if necessary. a. What is the probability that a randomly selected customer spends less than $85 at this store? b. What is the probability that a randomly selected customer spends between $65 and $85 at this store? c. What is the probability that a randomly selected customer spends more than $45 at this store? d. Find the dollar amount such that 75% of all customers spend no more than this amount. e. Find the dollar amount such that 80% of all customers spend at least this amount. f. Find two dollar amounts, equidistant from the mean, such that 90% of all customer purchases are between these values. 5th percentile 95th percentile

Answer 1

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:

$z=\dfrac{X-\mu}{\sigma}=\dfrac{85-75}{20}=\dfrac{10}{20}=0.5\\\\\\P(X$

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:

$z_1=\dfrac{X_1-\mu}{\sigma}=\dfrac{65-75}{20}=\dfrac{-10}{20}=-0.5\\\\\\z_2=\dfrac{X_2-\mu}{\sigma}=\dfrac{85-75}{20}=\dfrac{10}{20}=0.5\\\\\\\\P(65$

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.

$z=\dfrac{X-\mu}{\sigma}=\dfrac{45-75}{20}=\dfrac{-30}{20}=-1.5\\\\\\P(X>45)=P(z>-1.5)=0.933$

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:

$P(z$

Then, we can calculate X as:

$X^*=\mu+z^*\cdot\sigma=75+0.67449\cdot 20=75+13.4898=88.490$

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:

$P(z>-0.84162)=0.80$

Then, we can calculate X as:

$X^*=\mu+z^*\cdot\sigma=75+(-0.84162)\cdot 20=75-16.8324=58.168$

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:

$P(|z|$

The value for z* is ±1.64485.

We can now calculate the values for X as:

$X_1=\mu+z_1\cdot\sigma=75+(-1.64485)\cdot 20=75-32.897=42.103\\\\\\X_2=\mu+z_2\cdot\sigma=75+1.64485\cdot 20=75+32.897=107.897$

5th percentile: $42.103

95th percentile: $107.897