Answered

Birthweights at a local hospital have a Normal distribution with a mean of 110 oz and a standard deviation of 15 oz.


The proportion of infants with birthweights above 125 oz is

A0.500.
B0.159.
C0.341.
D0.841.

The proportion of infants with birthweights between 125 oz and 140 oz is

A0.819.
B0.636.
C0.477.
D0.136.

Verified Answer :

JeanaShupp

Answer:    B. 0.159.

D. 0.136.

Explanation:

Given : Birthweights at a local hospital have a Normal distribution with a mean of 110 oz and a standard deviation of 15 oz.

i.e. \(\mu=110\ ;\ \sigma=15\)

Formula to find z-score :-

\(z=\dfrac{x-\mu}{\sigma}\)

(a) To find the proportion of infants with birthweights above 125 oz, we find the z-score corresponds to 125 will be :-

\(z=\dfrac{125-110}{15}=1\)

By using the standard normal distribution table , The probability of infants with birthweights above 125 oz :-

\(P(x>125)=P(z>1)=1-P(z<1)=1- 0.8413447=0.1586553\approx0.159\)

Hence, the proportion of infants with birthweights above 125 oz is 0.159 .

(b) The z-score corresponds to 140 , \(z=\dfrac{140-110}{15}=2\)

The probability of infants with birthweights between 125 oz and 140 oz  :-

\(P(125

Hence, the proportion of infants with birthweights between 125 oz and 140 oz is 0.136.

fichoh

Using the normal distribution principle, the proportion of infants with birth weight above 125 and birth weight between 125 and 140 is ; 0.159 and 0.136 respectively

To obtain the Zscore :

  • Zscore = (X - μ) ÷ σ

A.)

Birth weight above 125 :

P(x > 125) = Z > (125 - 110) / 15))

P(x > 125) = Z > 1

Using a normal distribution calculator :

P(Z > 1) = 0.15866 = 0.159

B.)

Birth weight between 125 and 140:

P(x < 140) - P(X < 125) = Z < (140 - 110) / 15)) - Z > (125 - 110) / 15))

P(Z < 2) - P(Z < 1) = (0.97725 - 0.84134) = 0.1359 = 0.136

Therefore, the respective proportions are 0.159 and 0.136

Learn more :https://brainly.com/question/8165716

Related Questions