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3 years ago

Suppose a normal distribution has a mean of 98 and a standard deviation of 6. What is P(x< or = to 110)

Mathematics

1 answer:

dem82 [27]3 years ago

4 0

Answer:

97.8%

Step-by-step explanation:

110 is 2 standard deviations above the mean (6+6 = 12)

12+98 = 110

Looking at the standard deviation curve

P(x< or = to 110) = 1 - P(x>110)

We can find the probability that x>100 by adding anything above 2 standard deviations above the curve.

P(x>110) = 2.1+.1 = 2.2%

P(x< or = to 110) = 1 - P(x>110)

= 1- 2.2%

= 1- .022

= .978

= 97.8 %

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(Uniform) Suppose X follows a continuous uniform distribution from 4 to 11. (a) Write down the PDF of X. (b) Find P(X ≤ 7). Roun

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Answer:

a) $f(x)= \frac{1}{11-4}= \frac{1}{7}, 4 \leq x \leq 11$

b) $P(X\leq 7) = F(7) = \frac{7-4}{11-4}= 0.4286$

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d) $P(X >5 | X \leq 7)$

And we can find this probability with this formula from the Bayes theorem:

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Step-by-step explanation:

For this case we assume that the random variable X follows this distribution:

$X \sim Unif (a=4, b =11)$

Part a

The probability density function is given by the following expression:

$f(x) = \frac{1}{b-a} , a \leq x \leq b$

$f(x)= \frac{1}{11-4}= \frac{1}{7}, 4 \leq x \leq 11$

Part b

We want this probability:

$P(X \leq 7)$

And we can use the cumulative distribution function given by:

$F(x) = \frac{x-a}{b-a}= \frac{x-4}{11-4}$

And replacing we got:

$P(X\leq 7) = F(7) = \frac{7-4}{11-4}= 0.4286$

Part c

We want this probability:

$P(5 < X \leq 7)$

And we can use the CDF again and we have:

$P(5 < X \leq 7)= F(7) -F(5) = \frac{7-4}{7} -\frac{5-4}{7}= 0.2857$

Part d

We want this conditional probabilty:

$P(X >5 | X \leq 7)$

And we can find this probability with this formula from the Bayes theorem:

$P(X >5 | X \leq 7)= \frac{P(X>5 \cap X \leq 7)}{P(X \leq 7)}= \frac{P(5$

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