Question

The weights of newborn baby boys born at a local hospital are believed to have a normal distribution with a mean weight of 4004 grams and a variance of 103,684. If a newborn baby boy born at the local hospital is randomly selected, find the probability that the weight will be less than 4293 grams. Round your answer to four decimal places.

Answer 1

Answer:

0.8151 is the probability that the weight will be less than 4293 grams.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 4004 grams

Variance = 103,684

$\sigma = \sqrt{103684} = 322$

We are given that the distribution of weight of newborn baby is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

P(weight will be less than 4293 grams)

P(x < 4293)

$P( x < 4293) \\\\= P( z < \displaystyle\frac{4293 - 4004}{322}) \\\\= P(z < 0.8975)$

Calculation the value from standard normal z table, we have,  

$P(x < 4293) =0.8151 = 81.51\%$

0.8151 is the probability that the weight will be less than 4293 grams.

Answer 2

Answer:

Probability that the weight will be less than 4293 grams is 0.8133.

Step-by-step explanation:

We are given that weights of newborn baby boys born at a local hospital are believed to have a normal distribution with a mean weight of 4004 grams and a variance of 103,684.

Let X = weight of newborn baby boys

So, X ~ N($\mu =4004,\sigma^{2}=322^{2}$)

The z score probability distribution is given by;

               Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)  

where, $\mu$ = population mean

           $\sigma$ = population standard deviation

(a) Probability that weight will be less than 4293 grams is given by = P(X < 4293 grams)

    P(X < 4293) = P( $\frac{X-\mu}{\sigma}$ < $\frac{4293-4004}{322 }$ ) = P(Z < 0.89) = 0.8133

Therefore, if a newborn baby boy born at the local hospital is randomly selected, probability that the weight will be less than 4293 grams is 0.8133.