Question

I just need help with B.

In a study of a common insect, the number of bristles on the fifth abdominal sternite was shown to follow a normal distribution with mean 17.8 and standard deviation 1.7. (a) What percentage of the male population has fewer than 15 abdominal bristles? (b) Find an interval centered at the mean so that 90% of the male population have bristle numbers that fall into this interval. Click the icon to view the table of the standard normal distribution. X-μ (a) Use the transformation z= to find the value of z. 0 Z= - 1.65 (Round to two decimal places as needed.) Approximately 4.95 % of the male population has fewer than 15 abdominal bristles. (Round to two decimal places as needed.) (b) The interval is to abdominal bristles. (Round to one decimal place as needed.)

Answer 1

Only answer for part (b) has been asked.

(b)We need an interval centered around the mean which covers 90% of male population. Let this interval be (L, U). The area of the distribution outside this interval has 100%-90%=10% of the population. SInce normal distriburtion is symmetric, this means that the part of the distribution from  $\small -\infty$ to L has 5% of the population and part of the distribution from U to $\small \infty$ has 5% of the population. Area between L and U has 90% of the population. Therefore area between $\small -\infty$ and U has 5% + 90% = 95% of the population.

Let CDF be the cumulative density function of the Standard Normal distribution. Converting L and U to standard normal variables, we get Z1 = (L - 17.8)/1.7, Z2 = (U -17.8)/1.7

Hence CDF(Z1) = 0.05 and CDF(Z2) = 0.95.

Hence, Z1 = -1.645 and Z2 = 1.645

Or, (L - 17.8)/1.7 = -1.645 and (U - 17.8)/1.7 = 1.645

Or L = 15.0035 and U = 20.5965

Hence, the interval is 15.0 to 20.6 abdominal bristles.