Answer to Question #292288 in Statistics and Probability for Yaku

Question

18. The lengths of fully-grown scorpions of a certain variety have a mean of 1.96 inches and standard deviation of 0.08 inch. Assuming that the distribution of these lengths has roughly the shape of a normal distribution, find what percentage of these scorpions have a length of

a) 2.20 inches or more;

b) at least 1.80 inches.

19. With reference to Exercise 18, above what value would we find the longest 6 percent of these scorpions?

20. The distribution of the IQ's of the 4,000 employees of a large company has a mean of 104.5, a standard deviation of 13.9, and its shape is roughly that of a normal distribution. Given that a certain job requires a minimum IQ of 95 and bores those with an IQ over 110, how many of the company's employees are suitable for this job on the basis of IQ alone?

21. For a chi-square distribution, find

a) x²_0.025 when df = 15;

b) x²_0.01 when df = 7;

c) x²_0.05 when df = 24

Answer 1

18

Let $X$ be a random variable representing the length of full grown scorpions. $X\sim N(\mu=1.96,\sigma^2=0.08^2)$

$a)$

We find,

$p(x\ge 2.20)=p(Z\ge{2.20-\mu\over \sigma})=p(Z\ge{2.20-1.96\over 0.08})=p(Z\ge3)=1-p(Z\le3)=1-0.9987=0.0013$

In percentage, we have $0.0013\times100\%=0.13\%$ of the scorpions have length of 2.20 inches or more.

$b)$

We determine,

$p(x\ge 1.80)=p(Z\ge{1.80-\mu\over \sigma})=p(Z\ge{1.80-1.96\over 0.08})=p(Z\ge-2)=1-p(Z\le-2)=1-0.0228=0.9772$

In percent form, we have $0.9772\times100\%=97.72\%$ of the scorpions have length of at least 1.80 inches.

19

Finding the longest 6 percent of these scorpions is same as calculating the $94^{th}$ percentile.

To determine the value of the $94^{th}$ percentile, we find a value $y$ such that $p(x\lt y)=0.94$. Standardizing,

$p(x\lt y)=p(Z\lt {y-\mu\over \sigma})=p(Z\lt {y-1.96\over 0.08})=0.94$. To solve for the value of $y$, we determine the table value associated with 0.94 given as $Z_{0.94}$ and equate it to ${y-1.96\over0.08}$ . We can write this as,

${y-1.96\over0.08}=Z_{0.94}$. From the standard normal tables, $Z_{0.94}= 1.554774$

Therefore,

${y-1.96\over0.08}=1.554774\implies y=1.96+0.1244=2.0845$

Thus, above the value $y=2.0845$ inches is where we will find the longest 6 percent of these scorpions.

20

$\mu=104.5\\N=4000\\\sigma=13.9$

We find,

$p(95\lt x\lt 110)=p({95-104.5\over 13.9}\lt Z\lt{110-104.5\over 13.9})=p(-0.68\lt Z\lt 0.40)=\phi(0.40)-\phi(-0.68)=0.6554-0.2483=0.4071$

Having found the probability of the employees suitable for the job, we then multiply this probability by $N$ to find the number of employees suitable for this job.

So, the number of employees suitable for the job is $0.4071\times 4000=1628.4\approx 1629$.

Therefore, on the basis of IQ alone, the number of employees suitable for this job is approximately 1629 employees.

21

$a)$

To find $\chi^2_{0.025}$ when df = 15, we use the the chi-square tables.

From the tables, $\chi^2_{{0.025},15}=27.484$

$b)$

Same as part a above, $\chi^2_{{0.01},7}=18.4753$

$c)$

$\chi^2_{{0.05},24}=36.4151$ from chi-square tables.