High School

Assume that the newborn baby's weight in a hospital is normally distributed with a mean of 3.00 kg and a standard deviation of 0.25 kg.

(a) Find the probability that a baby born in the hospital has a weight between 2.75 kg and 3.20 kg. (Round your answer to 2 decimal places.)

(b) If 10 babies are picked randomly, find the probability that at least two babies have a weight of more than 3.25 kg.

(c) If 200 babies with weights between 2.75 kg and 3.20 kg are randomly selected, find the probability that between 116 and 135 of these babies have a weight within this range.

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]

  1. For 2.75 kg:
    [tex]Z_1 = \frac{2.75 - 3.00}{0.25} = -1.00[/tex]

  2. 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:

  1. Mean [tex]\mu_Y = np = 200 \times 0.6294 = 125.88[/tex]
  2. Variance [tex]\sigma_Y^2 = np(1-p) = 200 \times 0.6294 \times 0.3706 \approx 46.4336[/tex]
  3. 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.