Question

How much should a healthy kitten weigh? Suppose that a healthy 10-week-old (domestic) kitten should weigh an average of μ = 24.9 ounces with a (95% of data) range from 13.6 to 36.2 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 11% 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) it is given that range is from 36.2 to 13.6

Using rule of thumb, standard deviation = (upper limit - lower limit)/4 = (36.2-13.6)/4 = 5.65

Therefore, standard deviation = 5.65 and mean 24.9

(B) probability that a healthy 10-week-old kitten will weigh less than 14 ounces

using normalcdf(lower bound, upper bound, mean, standard deviation)

placing lower bound = negative infinity or -E99, upper bound = 14, mean = 24.9 and standard deviation = 5.65

this implies

= normalcdf(-E99, 14, 24.9,5.65)

= 0.0269

(C) probability that a healthy 10-week-old kitten will weigh more than 33 ounces

using normalcdf(lower bound, upper bound, mean, standard deviation)

placing lower bound = 33, upper bound = infinity or E99, mean = 24.9 and standard deviation = 5.65

this implies

= normalcdf(33,E99, 24.9,5.65)

= 0.0758

(D) probability that a healthy 10-week-old kitten will weigh between 14 and 33 ounces

using normalcdf(lower bound, upper bound, mean, standard deviation)

placing lower bound = 14, upper bound = 33, mean = 24.9 and standard deviation = 5.65

this implies

= normalcdf(14,33, 24.9,5.65)

= 0.8973

(E) Using InvNorm(z) to find the z corresponding to bottom 11%, we get

InvNorm(0.11) = -1.227 ...........................(using percentile table also, we get the same value)

Required cut off value = mean + (z*standard deviation)

= 24.9 + (-1.227*5.65)

= 24.9-6.933

= 17.93 (rounded to 2 decimals)