Question

sale amounts during lunch hour at a local subway are normally distributed, with a mean $7.27, and a standard deviation of $2.23.
a. find the probability that a randomly selected sale was at least $1.78? round answer to 4 decimal places.
b. a particular sale was $4.84. what is the percentile rank for this sale amount? round answer to the nearest percentage. hint: round proportion to two decimal places then convert to percent.
c. give the sale amount that is the cutoff for the highest 20 %? round answer to 2 decimal places.
d.what is the probability that a randomly selected sale is between $4.00 and $8.00? round answer to 4 decimal places.
e. what sale amount represents the cutoff for the middle 79 percent of sales? round answers to 2 decimal places. ( the smaller number here) (bigger number here)

Explanation:

Step1: Calculate z - score formula

The z - score is calculated as $z=\frac{x-\mu}{\sigma}$, where $\mu$ is the mean, $\sigma$ is the standard deviation, and $x$ is the value from the data set. Given $\mu = 7.27$ and $\sigma=2.23$.

Step2: Solve part a

For $x = 1.78$, the z - score is $z=\frac{1.78 - 7.27}{2.23}=\frac{- 5.49}{2.23}\approx - 2.46$. We want $P(X\geq1.78)$, which is equivalent to $P(Z\geq - 2.46)$. Since $P(Z\geq z)=1 - P(Z < z)$, and from the standard normal table $P(Z < - 2.46)=0.0069$, so $P(Z\geq - 2.46)=1 - 0.0069 = 0.9931$.

Step3: Solve part b

For $x = 4.84$, the z - score is $z=\frac{4.84 - 7.27}{2.23}=\frac{-2.43}{2.23}\approx - 1.09$. From the standard - normal table, $P(Z < - 1.09)=0.1379\approx0.14$. The percentile rank is $14\%$.

Step4: Solve part c

We want to find the value $x$ such that $P(X>x)=0.20$, or $P(X\leq x)=0.80$. Looking up the z - value in the standard - normal table for a probability of $0.80$, we get $z\approx0.84$. Then, using the z - score formula $z=\frac{x-\mu}{\sigma}$, we can solve for $x$: $0.84=\frac{x - 7.27}{2.23}$, so $x=0.84\times2.23 + 7.27=1.8732+7.27 = 9.1432\approx9.14$.

Step5: Solve part d

For $x_1 = 4.00$, the z - score is $z_1=\frac{4.00 - 7.27}{2.23}=\frac{-3.27}{2.23}\approx - 1.47$. For $x_2 = 8.00$, the z - score is $z_2=\frac{8.00 - 7.27}{2.23}=\frac{0.73}{2.23}\approx0.33$. Then $P(4.00

Step6: Solve part e

The middle $79\%$ of the data means that the area in the two tails is $1 - 0.79 = 0.21$, so the area in each tail is $\frac{0.21}{2}=0.105$. The z - value for the left - hand tail is $z_1$ such that $P(Z < z_1)=0.105$, and from the standard - normal table $z_1\approx - 1.25$. The z - value for the right - hand tail is $z_2$ such that $P(Z < z_2)=0.105 + 0.79=0.895$, and $z_2\approx1.25$.
For $z_1=-1.25$, using $z=\frac{x-\mu}{\sigma}$, we have $-1.25=\frac{x_1 - 7.27}{2.23}$, so $x_1=-1.25\times2.23 + 7.27=-2.7875 + 7.27 = 4.4825\approx4.48$.
For $z_2 = 1.25$, using $z=\frac{x-\mu}{\sigma}$, we have $1.25=\frac{x_2 - 7.27}{2.23}$, so $x_2=1.25\times2.23+7.27 = 2.7875+7.27 = 10.0575\approx10.06$.

Answer:

a. $0.9931$
b. $14\%$
c. $9.14$
d. $0.5585$
e. $4.48$, $10.06$