Question

Answer both questions all parts complete to get 100% feedback or dont attempt please!! Round off to nearest whole numbers.

The mean incubation time of fertilized eggs is 23 days. Suppose the incubation times are approximately normally distributed with a standard deviation of 1 day (a) Determine the 15th percentile for incubation times (6) Determine the incubation times that make up the middle 95% of fertilized eggs. (a) The 15th percentile for incubation times is days (Round to the nearest whole number as needed) (b) The incubation times that make up the middle 95% of fertilized eggs are todays, (Round to the nearest whole number as needed. Use ascending order)

Answer 1

1.

Suppose, random variable X denotes incubation time (in days) of fertilized eggs.

:: X~ N (u = 23,62 = 12)

(a)

We know,

$\tiny P\left ( Z<-1.036433 \right )=0.15$   [Using R-code 'qnorm(0.15)']

$\tiny \Rightarrow P\left ( \frac{X-\mu}{\sigma}<-1.036433 \right )=0.15$

$\tiny \Rightarrow P\left ( \frac{X-23}{1}<-1.036433 \right )=0.15$

= P(X < 23 – 1.036433 * 1) = 0.15

$\tiny \therefore P\left (X<21.96357 \right )=0.15$

Hence, the 15th percentile for incubation times is 22 days.

(b)

We know,

$\tiny P\left ( -1.959964<Z<1.959964 \right )=0.95$   [Using R-code 'qnorm(1-(1-0.95)/2)']

$\tiny \Rightarrow P\left ( -1.959964<\frac{X-\mu}{\sigma}<1.959964 \right )=0.95$

$\tiny \Rightarrow P\left ( -1.959964<\frac{X-23}{1}<1.959964 \right )=0.95$

$\tiny \Rightarrow P\left ( 23-1.959964<X<23+1.959964 \right )=0.95$

$\tiny \therefore P\left ( 21.04004<X<24.95996 \right )=0.95$

Hence, the incubation times that make up the middle 95% of fertilized eggs are 21 to 25 days.

2.

$\tiny E=\left \{ 11,12,13,14,15,16 \right \}$

$\tiny G=\left \{ 18,19,20,21 \right \}$

(a)

$\tiny E\cap G=\left \{ \right \}$

Hence, option C is correct.

(b)

$\tiny \because E\cap G=\left \{ \right \}$

So, E and G are mutually exclusive.

Hence, C. Yes, E and G have no outcome in common.