Question

In: Statistics and Probability

How much should a healthy kitten weigh? Suppose that a healthy 10-week-old (domestic) kitten should weigh...

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

Solutions

Expert Solution

(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)


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