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on a typical saturday, fiesta market reports that the mean amount of money spent by customers is $27.21 and a standard deviation of $7.93. the distribution is approximately normal.
- what percentage of people spend less than $20? 2) what percentage of people spend above $40?
- how many people spend between $25 and $30?
find the percentage:
lower/less than/below
what percent of students scored less than a score of x?
greater/more than/above
what percent of students scored above a score of x?
between
what percent of students scored between xlow and xhigh?
Step1: Calculate the z - score for $x = 20$
The formula for the z - score is $z=\frac{x-\mu}{\sigma}$, where $\mu = 27.21$ and $\sigma=7.93$.
Substitute $x = 20$ into the formula: $z_1=\frac{20 - 27.21}{7.93}=\frac{-7.21}{7.93}\approx - 0.91$.
Using a standard normal table or a calculator with a normal - distribution function (e.g., in Excel NORM.S.DIST(-0.91,TRUE)), we find $P(Z\lt - 0.91)\approx0.1814$ or $18.14\%$.
Step2: Calculate the z - score for $x = 40$
Substitute $x = 40$ into the z - score formula: $z_2=\frac{40 - 27.21}{7.93}=\frac{12.79}{7.93}\approx1.61$.
Using the property $P(Z\gt z)=1 - P(Z\lt z)$. For $z = 1.61$, using a standard normal table or NORM.S.DIST(1.61,TRUE) in Excel, $P(Z\lt1.61)\approx0.9463$. Then $P(Z\gt1.61)=1 - 0.9463 = 0.0537$ or $5.37\%$.
Step3: Calculate the z - scores for $x = 25$ and $x = 30$
For $x = 25$: $z_3=\frac{25 - 27.21}{7.93}=\frac{-2.21}{7.93}\approx - 0.28$.
For $x = 30$: $z_4=\frac{30 - 27.21}{7.93}=\frac{2.79}{7.93}\approx0.35$.
Using the formula $P(a\lt Z\lt b)=P(Z\lt b)-P(Z\lt a)$.
For $z=-0.28$, $P(Z\lt - 0.28)\approx0.3897$ (using NORM.S.DIST(-0.28,TRUE)).
For $z = 0.35$, $P(Z\lt0.35)\approx0.6368$ (using NORM.S.DIST(0.35,TRUE)).
Then $P(-0.28\lt Z\lt0.35)=0.6368 - 0.3897=0.2471$ or $24.71\%$.
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- Approximately $18.14\%$ of people spend less than $\$20$.
- Approximately $5.37\%$ of people spend above $\$40$.
- Approximately $24.71\%$ of people spend between $\$25$ and $\$30$.