Respuesta :
Answer:
Required Probability = 0.97062
Step-by-step explanation:
We are given that the weights of newborn baby boys born at a local hospital are believed to have a normal distribution with a mean weight of 4016 grams and a standard deviation of 532 grams.
Let X = weight of the newborn baby, so X ~ N([tex]\mu=4016 , \sigma^{2} = 532^{2}[/tex])
The standard normal z distribution is given by;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)
Now, probability that the weight will be less than 5026 grams = P(X < 5026)
P(X < 5026) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{5026-4016}{532}[/tex] ) = P(Z < 1.89) = 0.97062
Therefore, the probability that the weight will be less than 5026 grams is 0.97062 .
