Question

How much should a healthy kitten weigh? Suppose that a healthy 10-week-old (domestic) kitten should weigh an average of μ = 25.3 ounces with a (95% of data) range from 15.0 to 35.6 ounces. Let x be a random variable that represents the weight (in ounces) of a healthy 10-week-old kitten. Assume that x has a distribution that is approximately normal.

(a) The empirical rule (Section 7.1) indicates that for a symmetrical and bell-shaped distribution, approximately 95% of the data lies within two standard deviations of the mean. Therefore, a 95% range of data values extending from μ − 2σ to μ + 2σ is often used for "commonly occurring" data values. Note that the interval from μ − 2σ to μ + 2σ is 4σ in length. This leads to a "rule of thumb" for estimating the standard deviation from a 95% range of data values.

Estimating the standard deviation
For a symmetric, bell-shaped distribution,

STANDARD DEVIATION

range 4

≈

high value − low value 4

where it is estimated that about 95% of the commonly occurring data values fall into this range.

Estimate the standard deviation of the x distribution. (Round your answer to two decimal places.)
oz

(b) What is the probability that a healthy 10-week-old kitten will weigh less than 14 ounces? (Round your answer to four decimal places.)


(c) What is the probability that a healthy 10-week-old kitten will weigh more than 33 ounces? (Round your answer to four decimal places.)


(d) What is the probability that a healthy 10-week-old kitten will weigh between 14 and 33 ounces? (Round your answer to four decimal places.)


(e) A kitten whose weight is in the bottom 7% of the probability distribution of weights is called undernourished. What is the cutoff point for the weight of an undernourished kitten? (Round your answer to two decimal places.)
oz

Answer 1

Answer:

Population mean = 25.3

a)

95% data lies within 2 standard deviations from the mean

(35.6-25.3) or (25.3-15.0) = 2*sigma

=> Sigma = 10.3/2 = 5.15

Standard Deviation = 5.15

b)

P(X< 14) = P(Z< ((14-25.3) / 5.15)

= P(Z< -2.19 )

= 1 - P(Z<2.19)

= 1 - 0.9857

= 0.0143

c)

P(X > 33 ) = P(Z> ((33-25.3) / 5.15)

= P(Z > 1.5 )

= 1 - P(Z<1.5)

= 1 - 0.9332

= 0.0668

d) P( 14 < x < 33) =

1 - 0.0143 - 0.0668

= 0.9189

e)

C = 100 - 7 = 93%

Zc = Z 0.93 = -1.48

X = μ + 2σ

= 25.3 - (1.48 * 5.15) = 25.3 - 7.622 = 17.678 = 17.68 ounces