Answer :
Let's tackle the problem step-by-step using the properties of the normal distribution.
Part (a): Probability of a baby's weight between 2.75 kg and 3.20 kg
The weights of newborn babies are normally distributed with a mean [tex]\mu = 3.00[/tex] kg and a standard deviation [tex]\sigma = 0.25[/tex] kg.
We need to find the probability that a baby's weight [tex]X[/tex] lies between 2.75 kg and 3.20 kg. This can be found using the standard normal distribution (Z-distribution):
Convert the weights to Z-scores using the formula:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
For 2.75 kg:
[tex]Z_1 = \frac{2.75 - 3.00}{0.25} = -1.00[/tex]For 3.20 kg:
[tex]Z_2 = \frac{3.20 - 3.00}{0.25} = 0.80[/tex]
Now find the probability that [tex]-1.00 \leq Z \leq 0.80[/tex] using a standard normal distribution table or a calculator.
- Probability [tex]P(Z < 0.80) \approx 0.7881[/tex]
- Probability [tex]P(Z < -1.00) \approx 0.1587[/tex]
The probability that a baby has weight between 2.75 kg and 3.20 kg is then:
[tex]P(-1.00 \leq Z \leq 0.80) = P(Z < 0.80) - P(Z < -1.00) = 0.7881 - 0.1587 = 0.6294[/tex]
Rounded to two decimal places, the probability is 0.63.
Part (b): Probability that at least two out of 10 babies weigh more than 3.25 kg
First, calculate the probability that a single baby weighs more than 3.25 kg:
Convert 3.25 kg to a Z-score:
[tex]Z = \frac{3.25 - 3.00}{0.25} = 1.00[/tex]
Find the probability:
- Probability [tex]P(Z > 1.00) \approx 1 - P(Z < 1.00) \approx 1 - 0.8413 = 0.1587[/tex]
We have a binomial distribution problem here where each baby's weight being more than 3.25 kg is a "success" with probability [tex]p = 0.1587[/tex]. If there are 10 babies (trials), we need to find the probability that at least 2 babies have this weight.
This is given by:
[tex]P(X \geq 2) = 1 - P(X = 0) - P(X = 1)[/tex]
Where [tex]X[/tex] is a binomial random variable with [tex]n = 10[/tex] and [tex]p = 0.1587[/tex].
- [tex]P(X = 0) = \binom{10}{0} (0.1587)^0 (0.8413)^{10} \approx 0.2096[/tex]
- [tex]P(X = 1) = \binom{10}{1} (0.1587)^1 (0.8413)^9 \approx 0.3954[/tex]
Therefore,
[tex]P(X \geq 2) = 1 - 0.2096 - 0.3954 \approx 0.3950[/tex]
Part (c): Probability that 116 to 135 out of 200 babies weigh between 2.75 kg and 3.20 kg
From part (a), we found that approximately 63% of babies fall within the 2.75 to 3.20 kg range. Let [tex]Y[/tex] be the number of babies in this range from a sample of 200.
[tex]Y \sim \text{Binomial}(200, 0.6294)[/tex]
Finding the probability [tex]P(116 \leq Y \leq 135)[/tex] can be complex by hand, but can be approximated using a normal distribution:
- Mean [tex]\mu_Y = np = 200 \times 0.6294 = 125.88[/tex]
- Variance [tex]\sigma_Y^2 = np(1-p) = 200 \times 0.6294 \times 0.3706 \approx 46.4336[/tex]
- Standard deviation [tex]\sigma_Y \approx \sqrt{46.4336} \approx 6.8167[/tex]
Using the normal approximation:
Convert 116 and 135 to Z-scores:
[tex]Z_{116} = \frac{116 - 125.88}{6.8167} \approx -1.449[/tex]
[tex]Z_{135} = \frac{135 - 125.88}{6.8167} \approx 1.340[/tex]
Using standard normal distribution tables or a calculator:
- Probability [tex]P(Z < 1.340) \approx 0.9099[/tex]
- Probability [tex]P(Z < -1.449) \approx 0.0730[/tex]
So,
[tex]P(116 \leq Y \leq 135) = P(Z < 1.340) - P(Z < -1.449) \approx 0.9099 - 0.0730 = 0.8369[/tex]
The probability is approximately 0.84.
