Question

1. door dash has a policy allowing customers to cancel an order after waiting an hour (60 minutes). delivery times are known to be normally distributed with a mean of 35 minutes and a standard deviation of 10 minutes. what proportion of orders would be eligible for cancellation? a.).0062 c.).6767 b.).1214 d.).7676 13. door dash has a policy allowing customers to cancel an order after waiting an hour (60 minutes). delivery times are known to be normally distributed with a mean of 35 minutes and a standard deviation of 10 minutes. what delivery time has a z score of -1? a.) 10 minutes c.) 35 minutes b.) 25 minutes d.) 67 minutes 14. door dash has a policy allowing customers to cancel an order after waiting an hour (60 minutes). delivery times are known to be normally distributed with a mean of 35 minutes and a standard deviation of 10 minutes. what proportion of orders take less than 20 minutes? a.).6771 c.).0668 b.).3229 d.).9332 15. door dash has a policy allowing customers to cancel an order after waiting an hour (60 minutes). delivery times are known to be normally distributed with a mean of 35 minutes and a standard deviation of 10 minutes. what proportion of orders take more than 50 minutes? a.).0067 c.).9332 b.).0668 d.).6771 16. door dash has a policy allowing customers to cancel an order after waiting an hour (60 minutes). delivery times are known to be normally distributed with a mean of 35 minutes and a standard deviation of 10 minutes. how fast must an order be to be in the 5th percentile of delivery times? a.) 67 minutes c.) 18.55 minutes b.) 26.7 minutes d.) 10.36 minutes

Explanation:

Step1: Calculate z - score for 60 minutes

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(x = 60\), \(\mu=35\), and \(\sigma = 10\).

$$z=\frac{60 - 35}{10}=\frac{25}{10}=2.5$$

Step2: Find the proportion using the standard normal table

We want \(P(X\geq60)\), which is equivalent to \(P(Z\geq2.5)\).
Since \(P(Z\geq z)=1 - P(Z < z)\), and from the standard normal table \(P(Z < 2.5)=0.9938\)

$$P(Z\geq2.5)=1 - 0.9938=0.0062$$

Step3: Calculate z - score for 20 minutes

Using \(z=\frac{x-\mu}{\sigma}\), with \(x = 20\), \(\mu = 35\), \(\sigma=10\)

$$z=\frac{20 - 35}{10}=\frac{- 15}{10}=-1.5$$

Step4: Find the proportion for \(x < 20\)

Using the standard normal table, \(P(Z < - 1.5)=0.0668\)

Step5: Calculate z - score for 50 minutes

Using \(z=\frac{x-\mu}{\sigma}\), with \(x = 50\), \(\mu = 35\), \(\sigma=10\)

$$z=\frac{50 - 35}{10}=\frac{15}{10}=1.5$$

Step6: Find the proportion for \(x>50\)

\(P(X > 50)=P(Z>1.5)\)
Since \(P(Z>z)=1 - P(Z\leq z)\) and \(P(Z\leq1.5)=0.9332\)

$$P(Z > 1.5)=1 - 0.9332=0.0668$$

Step7: Find the value for the 5th percentile

If \(P(X < x)=0.05\), then we first find the z - score corresponding to \(p = 0.05\). From the standard normal table, \(z=-1.645\)
Using the formula \(z=\frac{x-\mu}{\sigma}\), we solve for \(x\):

$$x=\mu+z\sigma$$

Substitute \(\mu = 35\), \(z=-1.645\), \(\sigma = 10\)

$$x=35+(-1.645)\times10=35 - 16.45=18.55$$

Answer:

1. a. \(0.0062\)
2. b. \(25\) minutes (since \(x=\mu+z\sigma=35+(-1)\times10 = 25\))
3. c. \(0.0668\)
4. b. \(0.0668\)
5. c. \(18.55\) minutes