Question

Please solve this question

Answer 1

The probability that a randomly selected x-value from the distribution will be in the interval:

• P(35 < x < 45) = 0.6827 and,
• P(30 < x < 40) = 0.47725

What is the probability of a normal distribution?

The probability of a normal distribution can be determined from the symmetrical curve between 1 to 100%.

From the information given:

• Mean = 40
• Standard deviation = 5

To determine the probability that a randomly selected x-value is in the given interval:

• P(35<x<45)
• P(30<x<40)

$P(35 < x < 45) = P(\dfrac{35-40}{5} < Z < \dfrac{45-40}{5})$

$P(35 < x < 45) = P(\dfrac{-5}{5} < Z < \dfrac{5}{5})$

$P(35 < x < 45) = P(-1 < Z < 1)$

$\mathbf{P(35 < x < 45) = P[Z\le 1] -P[Z\le -1]}$

Using normal distribution table:

P(35 < x < 45) = 0.8414 - 0.1587

P(35 < x < 45) = 0.6827

$P(30 < x < 40) = P(\dfrac{30-40}{5} < Z < \dfrac{40-40}{5})$

$P(30 < x < 40) = P(\dfrac{-10}{5} < Z < \dfrac{0}{5})$

$P(30 < x < 40) = P(-2 < Z < 0)$

$\mathbf{P(30 < x < 40) = P[Z\le0]-P[Z\le -2]}$

Using normal distribution table:

P(30 < x < 40) = 0.5 - 0.02275

P(30 < x < 40) = 0.47725

Learn more about the probability of a normal distribution here:

https://brainly.com/question/4079902

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