Question

If 25 new born babies are randomly selected, what is the probability that the 25 babies are born that their mean weigh less than 3100g?

Answer 1

a) The parameters for the sampling distribution include

Mean = μₓ = 3500 g

Standard deviation = σₓ = 100 g

Required probability = P(x < 3100)

b) The image of the probability density curve is presented in the attached image.

c) The z-score for 3100 g in the sampling distribution = -4.00

d) The probability that the 25 babies that are born that their mean weigh less than 3100 g = 0.00003

Explanation:

Complete Question

Suppose we know the birth weights of babies is normally distributed, with mean of 3500 g and standard deviation 500g.

If 25 new born babies are randomly selected, what is the probability that the 25 babies are born that their mean weigh less than 3100g?

a) What are the parameters?

b) Draw a probability density curve for the problem.

c) What is the z-score of the weight?

d) What is the required probability?

Solution

The Central limit theorem explains that for a random sample of adequate size obtained from a normal distribution with independent variables, the mean of the sampling distribution is approximately equal to the population mean and the sampling distribution is approximately of the nature of the population distribution (normal) with the standard deviation of the sampling distribution given as

Standard deviation of the sampling distribution = [(Population Standard deviation)/√n]

σₓ = (σ/√n)

Mean of Sampling distribution = Population mean

μₓ = μ = 3500 g

σₓ = (σ/√n) = (500/√25) = 100 g

a) The parameters for the sampling distribution include

Mean = μₓ = 3500 g

Standard deviation = σₓ = 100 g

Required probability = P(x < 3100)

b) The image of the probability density curve is presented in the attached image.

c and d) To obtain the required probability, we need the z-score for 3100 g, so we standardize 3100g

The standardized score for any value is the value minus the mean then divided by the standard deviation.

z = (x - μ)/σ = (3100 - 3500)/100 = - 4.00

To determine the required probability 45mg/L, P(x < 3100) = P(z < -4.00)

We'll use data from the normal probability table for these probabilities

P(x < 3100) = P(z < -4.00) = 0.00003

Hence, the probability that the 25 babies that are born that their mean weigh less than 3100 g = 0.00003

Hope this Helps!!!!

Answer 2

Explanation:

Given that

n = 25

mean = 3100

standard deviation = 500

`\bar x = (500)/(√(25) )`

= 500 / 5

= 100

N ( μ = 3500, (500)² / 25)

N = (3500,(100)² )

b) P ( x < 3100)

c) Z score

`z = \frac{x - u _(\bar x)}{\sigma \_ {\bar x}}`

`u_(\bar x)= 3500`

`\sigma _(\bar x)= 100`

`z = (3100-3500)/(100) \\\\= -4`

d)

probability=

P ( x < 3100) = P < -4 = 0