The birth weight of babies is normally distributed with mean 3500g 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? Find the z-score, and construct the standard normal distribution density curve, then shade your seeking area. Find the probability
Answers
Answer:
Answer:
a) The parameters for the sampling distribution include
Mean = μₓ = 3500 g
Standard deviation = σₓ = 100 g
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)
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