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question 4 test scores use the empirical rule to answer the following q…

Question

question 4

test scores

use the empirical rule to answer the following questions.

a. the mean is

b. the standard deviation is

c. % of the test scores are between 71 and 79.

d. % of the test scores are between 67 and 83.

e. % of the test scores are between 63 and 87.

f. % of the test scores are between 75 and 79.

g. % of the test scores are less than 75.

h. % of the test scores are less than 79.

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Explanation:

Step1: Find the mean

The normal distribution is symmetric about the mean. The mid - point of the distribution (the peak of the bell - curve) is at \(x = 75\). So, the mean \(\mu=75\).

Step2: Find the standard deviation

The distance between the mean and the inflection point (where the curve changes from concave up to concave down) is one standard deviation. The distance from \(75\) to \(71\) (or \(75\) to \(79\)) is \(4\). So, the standard deviation \(\sigma = 4\).

Step3: Use the Empirical Rule for part c

The Empirical Rule states that for a normal distribution:

  • \(P(\mu-\sigma<X<\mu + \sigma)\approx68\%\), \(P(\mu - 2\sigma<X<\mu+2\sigma)\approx95\%\), \(P(\mu-3\sigma<X<\mu + 3\sigma)\approx99.7\%\)

For \(71 = 75-4=\mu-\sigma\) and \(79=75 + 4=\mu+\sigma\), so \(P(71<X<79)\approx68\%\)

Step4: Use the Empirical Rule for part d

For \(67=75 - 2\times4=\mu-2\sigma\) and \(83=75+2\times4=\mu + 2\sigma\), so \(P(67<X<83)\approx95\%\)

Step5: Use the Empirical Rule for part e

For \(63=75-3\times4=\mu - 3\sigma\) and \(87=75+3\times4=\mu+3\sigma\), so \(P(63<X<87)\approx99.7\%\)

Step6: Use the Empirical Rule for part f

Since \(P(71<X<79)\approx68\%\), and the normal distribution is symmetric about the mean \(\mu = 75\). The interval \(75\) to \(79\) is half of the interval \(71\) to \(79\). So \(P(75<X<79)=\frac{68\%}{2}=34\%\)

Step7: Use the property of the normal distribution for part g

Since the normal distribution is symmetric about the mean \(\mu = 75\), \(P(X<75)=50\%\)

Step8: Use the Empirical Rule for part h

We know \(P(71<X<79)\approx68\%\) and \(P(X<71)=\frac{100 - 68}{2}\%=16\%\)
\(P(X<79)=P(X<71)+P(71<X<79)=16\%+68\% = 84\%\)

Answer:

a. \(75\)
b. \(4\)
c. \(68\)
d. \(95\)
e. \(99.7\)
f. \(34\)
g. \(50\)
h. \(84\)