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7 scientists measured the lengths of a species of bird in a region. the…

Question

7

scientists measured the lengths of a species of bird in a region. the graph below shows that the measured lengths have a normal distribution. about what percent of the bird population is between 14 and 34 cm long?
a) 78% b) 96%
c) 68% d) 88%

  1. a candy store owner wants to know the overall thoughts of customers in her store. she asks the first 20 customers who come into her store on a monday morning and ask them to rate there store experience on a scale of 1 to 10. her customers average response was a 8.8 and she concludes that all her customers have positive thoughts about her store. her conclusion is,

a) invalid, since the customers were chosen by b) unfair because she owns the store
convenience
c) spot on! d) valid, since customers were chosen randomly

Explanation:

Step1: Recall the empirical rule for normal distribution

The empirical rule states that for a normal distribution:

  • Approximately \(68\%\) of the data lies within \(1\) standard deviation (\(\sigma\)) of the mean (\(\mu\))
  • Approximately \(95\%\) of the data lies within \(2\) standard deviations (\(\sigma\)) of the mean (\(\mu\))
  • Approximately \(99.7\%\) of the data lies within \(3\) standard deviations (\(\sigma\)) of the mean (\(\mu\))

Let the mean \(\mu = 22\) (from the center of the normal - distribution graph). If \(x_1=14\) and \(x_2 = 34\), then \(\mu-\sigma=22 - 8=14\) and \(\mu + 3\sigma=22+12 = 34\) is incorrect. Wait, no, if we assume the distance from the mean to \(14\) is \(22-14 = 8\) and from \(22\) to \(34\) is \(34 - 22=12\) (wrong approach). Wait, actually, in a normal distribution, the total area under the curve is \(1\) (or \(100\%\)). The distance from \(14\) to \(34\):
The mean is \(22\). The value \(14=22 - 8\) and \(34=22+12\) (wrong). Wait, no, if we consider the symmetry of the normal distribution. The general formula for the percentage of data between \(a\) and \(b\) in a normal distribution \(N(\mu,\sigma)\) is \(P(a<X < b)=\Phi(\frac{b - \mu}{\sigma})-\Phi(\frac{a - \mu}{\sigma})\), but using the empirical rule (approximate):
If we assume that the distance from \(14\) to \(34\) covers \(3\) standard - deviation intervals. The mean is \(22\). The distance from \(14\) to \(22\) is \(8\) and from \(22\) to \(34\) is \(12\) (wrong). Wait, no, actually, if we consider that in a normal distribution:
The percentage of data within \(3\) standard deviations of the mean is about \(99.7\%\). But if we calculate the number of standard - deviation intervals:
Let’s assume the standard deviation \(\sigma = 4\) (since \(22-14 = 8\) ( \(2\sigma\)) and \(34 - 22=12\) ( \(3\sigma\)) is wrong. Wait, no, the correct way is:
The percentage of data within \(k\) standard deviations of the mean in a normal distribution is given by \(P(\mu - k\sigma<X<\mu + k\sigma)\).
If \(\mu = 22\), \(\mu-\sigma=18\), \(\mu - 2\sigma=14\), \(\mu+2\sigma = 30\), \(\mu+3\sigma=34\)
\(P(14<X<34)=P(\mu - 2\sigma<X<\mu+3\sigma)\)
\(P(\mu - 2\sigma<X<\mu)=47.5\%\) (from the empirical rule \(P(\mu - 2\sigma<X<\mu + 2\sigma)=95\%\), so \(P(\mu - 2\sigma<X<\mu)=47.5\%\)) and \(P(\mu<X<\mu + 3\sigma)=49.85\%\) (since \(P(\mu - 3\sigma<X<\mu + 3\sigma)=99.7\%\), so \(P(\mu<X<\mu + 3\sigma)=49.85\%\))
\(P(14<X<34)=47.5\%+49.85\%=97.35\%\approx98\%\)

Answer:

b) \(98\%\)