Question

How much should a healthy kitten weigh? Suppose that a healthy 10-week-old (domestic) kitten should weigh an average of μ = 24.7 ounces with a (95% of data) range from 14.0 to 35.4ounces. 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 9% 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

(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

Answer : Here standard deviation = Range/ 4 = (35.4 - 14.0)/4 = 5.35

(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.)

ANswer : Pr(x < 14 ounces) = NORM( x < 14 ounces ; 24.7 ounces ; 5.35 ounces)

Z = (14 - 24.7)/5.35 = -2

Pr(x < 14 ounces) = NORM( x < 14 ounces ; 24.7 ounces ; 5.35 ounces) =Pr(Z < -2) = 0.02275

(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.)

ANswer : Pr(x 33 ounces) = 1 - Pr(x < 33 ounces) = 1 - NORM( x < 33 ounces ; 24.7 ounces ; 5.35 ounces)

Z = (33 - 24.7)/5.35 = 1.5514

Pr(x 33 ounces) = 1 - Pr(x < 33 ounces) = 1 - NORM( x < 33 ounces ; 24.7 ounces ; 5.35 ounces) = 1 - Pr(Z < 1.5514) = 1 - 0.9396 = 0.0604

(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.)

= Pr(14 ounces < x < 33 ounces) = NORM( x < 33 ounces ; 24.7 ounces ; 5.35 ounces) - NORM( x < 14 ounces ; 24.7 ounces ; 5.35 ounces) = Pr(Z < 1.5514) - Pr(Z < -2)

= 0.9396 - 0.0228 = 0.9168

(e) A kitten whose weight is in the bottom 9% 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.)

Answer : here lets say the cutoff point is x₀

Pr(x < x₀ ) = 0.09 = NORMDIST(x < x₀ ; 24.7 oz, 5.35 oz)

Here checking the z table

Z = -1.34076 = (x₀ - 24.7)/5.35

x₀ = 24.7 - 5.35 * 1.34076 = 17.53 oz