How much should a healthy kitten weigh? Suppose that a healthy 10weekold (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 10weekold kitten. Assume that x has a distribution that is approximately normal.
(a) The empirical rule (Section 7.1) indicates that for a symmetrical and bellshaped 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, bellshaped distribution,
STANDARD DEVIATION

≈ 

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 10weekold kitten will
weigh less than 14 ounces? (Round your answer to four decimal
places.)
(c) What is the probability that a healthy 10weekold kitten will
weigh more than 33 ounces? (Round your answer to four decimal
places.)
(d) What is the probability that a healthy 10weekold 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:
Population mean = 25.3
a)
95% data lies within 2 standard deviations from the mean
(35.625.3) or (25.315.0) = 2*sigma
=> Sigma = 10.3/2 = 5.15
Standard Deviation = 5.15
b)
P(X< 14) = P(Z< ((1425.3) / 5.15)
= P(Z< 2.19 )
= 1  P(Z<2.19)
= 1  0.9857
= 0.0143
c)
P(X > 33 ) = P(Z> ((3325.3) / 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
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