Question

15. The grapefruits grown in a large orchard have a mean weight of 18.2 ounces with a standard deviation of 1.2 ounces. weights of these grapefruits has roughly the shape of a normal distribution, what percentage of the grapefruit weigh

a) less than 16.1 ounces;

b) more than 17.3 ounces;

c) anywhere from 16.7 to 18.8 ounces?

16. With reference to Exercise 15, find

a) the weight above which we will find the heaviest 15% and 75% of the grapefruits;

17. A manufacturer needs coil springs that can stand a load of at least 20.0 pounds. Among two suppliers, Supplier A can supply coil springs that, on the average, can stand a load of 24.5 pounds with a standard deviation of 2.1 pounds, and Supplier B can supply coil springs that, on the average, can stand a load of 23.3 pounds with a standard deviation of 1.6 pounds. distributions of these loads can be apprx with a normal distributions, determine which of the two suppliers can provide the manufacturer with the smaller percentage of unsatisfactory coil springs.

Answer 1

$\mu=18.2\\\sigma=1.2$

$15$

$a)$

Here, we find $p(x\lt16.1)=p(Z\lt{16.1-\mu\over\sigma})=p(Z\lt{16.1-18.2\over1.2})=p(Z\lt-1.75)=0.0401$

$b)$

$p(x\gt17.3)=p(Z\gt{17.3-\mu\over\sigma})=p(Z\gt{17.3-18.2\over1.2})=p(Z\gt-0.75)=1-p(Z\lt -0.75)=1-0.2266=0.7734$

$c)$

$p(16.7\lt x\lt 18.8)=p({16.7-18.2\over 1.2}\lt Z\lt {18.8-18.2\over1.2})=p(-1.25\lt Z\lt 0.5)=\phi(0.5)-\phi (-1.25)=0.6915-0.1056=0.5859$

16

To find the heaviest 15% is same as finding the $85^{th}$ percentile. Here, we determine a value $y$ such that, $p(x\lt y)=0.85\implies p(Z\lt {y-18.2\over 1.2})=0.85$. To find the value of $y$, we determine the standard normal table value associated with 0.85 and equate it to ${y-18.2\over 1.2}$ . That is, ${y-18.2\over 1.2}=Z_{0.85}=1.036433$. Solving for $y,$ we have,

${y-18.2\over 1.2}=1.036433\implies y=18.2+1.2437196=19.44$. Therefore, the weight above which we will find the heaviest 15% of the grapefruits is 19.44 ounces.

To find the heaviest 75% is same as finding the $25^{th}$ percentile. Here, we determine a value $t$ such that, $p(x\lt t)=0.25\implies p(Z\lt {t-18.2\over 1.2})=0.25$. To find the value of $t$, we determine the standard normal table value associated with 0.25 and equate it to ${t-18.2\over 1.2}$ . That is, ${t-18.2\over 1.2}=Z_{0.25}= -0.6744898$. Solving for $t,$ we have,

${t-18.2\over 1.2}= -0.6744898\implies t=18.2-0.8094=17.39$.

Therefore, the weight above which we will find the heaviest 75% of the grapefruits is 17.39 ounces.

17

Supplier A

$\mu=24.5\\\sigma=2.1$

We find the probability,

$p(X\lt 20)=p(Z\lt {20-\mu\over\sigma})=p(Z\lt{20-24.5\over2.1})=p(Z\lt-2.14)=0.0162$.

Therefore, the percentage of unsatisfactory coil springs is $0.0162\times100\%=1.62\%$

Supplier B

$\mu=23.3\\\sigma=1.6$

We find the probability,

$p(X\lt 20)=p(Z\lt {20-\mu\over\sigma})=p(Z\lt{20-23.3\over1.6})=p(Z\lt-2.06)=0.0197$.

Therefore, the percentage of unsatisfactory coil springs is $0.0197\times100\%=1.97\%$

Of the two suppliers, supplier A can supply a smaller percentage of unsatisfactory spring coils to the manufacturer.