Question

Suppose that the probability distribution for birth weights is normal with a mean of 120 ounces and a standard deviation of 20 ounces. The probability that a randomly selected infant has a birth weight between 100 ounces and 140 ounces is [ Select ] 68%. The probability that a randomly selected infant has a birth weight between 110 and 130 is [ Select ] 68%.

Answer 1

The probability that a randomly selected infant has a birth weight between 100 ounces and 140 ounces is 68%.

The probability that a randomly selected infant has a birth weight between 110 and 130 is 38%.

Explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the z-score of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Mean of 120 ounces and a standard deviation of 20 ounces.

This means that
`\mu = 120, \sigma = 20`

The probability that a randomly selected infant has a birth weight between 100 ounces and 140 ounces is

p-value of Z when X = 140 subtracted by the p-value of Z when X = 100.

X = 140

`Z = (X - \mu)/(\sigma)`

`Z = (140 - 120)/(20)`

`Z = 1`

`Z = 1` has a p-value of 0.84

X = 100

`Z = (X - \mu)/(\sigma)`

`Z = (100 - 120)/(20)`

`Z = -1`

`Z = -1` has a p-value of 0.16

0.84 - 0.16 = 0.68

The probability that a randomly selected infant has a birth weight between 100 ounces and 140 ounces is 68%.

The probability that a randomly selected infant has a birth weight between 110 and 130

This is the p-value of Z when X = 130 subtracted by the p-value of Z when X = 110.

X = 130

`Z = (X - \mu)/(\sigma)`

`Z = (130 - 120)/(20)`

`Z = 0.5`

`Z = 0.5` has a p-value of 0.69

X = 110

`Z = (X - \mu)/(\sigma)`

`Z = (110 - 120)/(20)`

`Z = -0.5`

`Z = -0.5` has a p-value of 0.31

0.69 - 0.31 = 0.38 = 38%.

The probability that a randomly selected infant has a birth weight between 110 and 130 is 38%.