Question

The gestation period for cats has an approximate mean of 64 days and a standard deviation of 3 days. The distribution of the gestation period is approximately Normal.
a. What proportion of kittens have a gestation period of less than 68 days?
b. What proportion of kittens have a gestation period between 61 and 70 days?
c. What gestation period corresponds to the top 10% of gestation periods?
d. What gestation period corresponds to the 25th percentile?

Answer 1

Answer:

a. P(X<68) = 0.9082

b. P(61 < X< 70) = 0.8185

c. the gestation period that corresponds to the top 10% of gestation periods = 67.846

d. the gestation period that corresponds to the 25th percentile = 61.977

Explanation:

Given that:

population mean  $\mu$ = 64

standard deviation $\sigma$ = 3

a. What proportion of kittens have a gestation period of less than 68 days?

here the sample mean x = 68

The standard normal distribution for the z score is

$z = \dfrac{x -\mu}{\sigma}$

$z = \dfrac{68 -64}{3}$

$z = \dfrac{4}{3}$

z = 1.33

The proportion of the kittens having a gestation period of less than 68 days is:

P(X<68) = P(Z< 1.33)

Using the z - tables

P(X<68) = 0.9082

b. What proportion of kittens have a gestation period between 61 and 70 days?

here ; sample mean x₁ = 61 and x₂ = 70

the standard normal distribution for the z score is:

$z_1 = \dfrac{61 -64}{3}$

$z_1 = \dfrac{-3}{3}$

$z_1 =-1$

$z_2 = \dfrac{70-64}{3}$

$z_2= \dfrac{6}{3}$

$z_2= 2$

So, the proportion of kittens having a gestation period between 61 and 70 days is:

P(61 < X< 70) = P(-1 < Z < 2)

P(61 < X< 70) = P(Z < 2) - P( Z< -1)

From z tables

P(61 < X< 70) = 0.9772 - 0.1587

P(61 < X< 70) = 0.8185

c.  What gestation period corresponds to the top 10% of gestation periods?

i.e

P(X >$x_o$ ) = 0.1

P(X < $x_o$ ) = 1 - 0.1

P(X >$x_o$ ) = 0.9

$P(Z < \dfrac{x - \mu}{\sigma}) =0.9$

Using the Excel Function : =NORMINV (0.9)

$P(Z < \dfrac{x - \mu}{\sigma}) =1.282$

⇒ $\dfrac{x - \mu}{\sigma}=1.282$

${x - \mu}=1.282 \times \sigma$

$x =1.282 \times \sigma + \mu$

given that:

$\mu = 64 \\ \sigma =3$

x = 1.282 × 3 + 64

x = 67.846

The gestation period that corresponds to the top 10% of gestation periods = 67.846

d. What gestation period corresponds to the 25th percentile?

At 25 percentile, using the EXCEL FUNCTION = NORMINV(0.25;64;3)

the gestation period that corresponds to the 25th percentile = 61.977