Question

The length of human pregnancies from conception to birth follows a distribution with mean 266 days and standard deviation 15 days.
1- Assume the distribution is bell-shaped (symmetric). The percent of pregnancies last between 236 and 281 days is approximately, [ Select ] ["81.5 %", "19.5%", "68%", "99.7%", "95%"]

2- - Assume the distribution is bell-shaped (symmetric). The percent of pregnancies last between 236 and 296 days is approximately, [ Select ] ["75%", "68%", "99.7%", "95%"]

3- - Assume the distribution is not bell-shaped ( non symmetric). The percent of pregnancies last between 236 and 296 days is approximately, [ Select ] ["85%", "75%", "99.7%", "88.9%", "95%"]

Answer 1

Answer:

a) 81.5%

b) 95%

c) 75%

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 266 days

Standard Deviation, σ = 15 days

We are given that the distribution of  length of human pregnancies is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) P(between 236 and 281 days)

$P(236 \leq x \leq 281)\\\\= P(\displaystyle\frac{236 - 266}{15} \leq z \leq \displaystyle\frac{281-266}{15})\\\\= P(-2 \leq z \leq 1)\\\\= P(z \leq 1) - P(z < -2)\\= 0.838 - 0.023 = 0.815 = 81.5\%$

b) a) P(last between 236 and 296)

$P(236 \leq x \leq 281)\\\\= P(\displaystyle\frac{236 - 266}{15} \leq z \leq \displaystyle\frac{296-266}{15})\\\\= P(-2 \leq z \leq 2)\\\\= P(z \leq 2) - P(z < -2)\\= 0.973 - 0.023 = 0.95 = 95\%$

c) If the data is not normally distributed.

Then, according to Chebyshev's theorem, at least $1-\dfrac{1}{k^2}$  data lies within k standard deviation of mean.

For k = 2

$1-\dfrac{1}{(2)^2} = 75\%$

Atleast 75% of data lies within two standard deviation for a non normal data.

Thus, atleast 75% of pregnancies last between 236 and 296 days approximately.