Question

You are planning a May camping trip to Denali National Park in Alaska and want to make sure your sleeping bag is warm enough. The average low temperature in the park for May follows a normal distribution with a mean of 32°F and a standard deviation of 8°F. One sleeping bag you are considering advertises that it is good for temperatures down to 25°F. What is the probability that this bag will be warm enough on a randomly selected May night at the park?
A 0.1894
B 0.3106C 0.8106D 0.8800

Answer 1

The probability that this bag will be warm enough on a randomly

selected May night at the park is 0.8106 ⇒ answer C

Step-by-step explanation:

You are planning a May camping trip to Denali National Park in Alaska

and want to make sure your sleeping bag is warm enough.

The average low temperature in the park for May follows a normal

distribution

The given is:

1. The mean is 32°F

2. The standard deviation is 8°F

3. One sleeping bag you are considering advertises that it is good for

   temperatures down to 25°F (x ≥ 25)

At first let us find the z-score

∵ z = (x - μ)/σ, where x is the score, μ is the mean, σ is standard deviation

∵ μ = 32°F , σ = 8°F , x = 25°F

- Substitute these values in the rule

∴ z = $\frac{25-32}{8}=-0.875$

Now let us find the corresponding area of z-score in the normal

distribution table

∵ The corresponding area of z = -0.875 is 0.18943

∵ For P(x ≥ 25) the area to the right is needed

∵ P(x ≥ 25) = 1 - 0.18943 = 0.8106

∴ P(x ≥ 25) = 0.8106

The probability that this bag will be warm enough on a randomly

selected May night at the park is 0.8106

Learn more:

You can learn more about mean and standard deviation in brainly.com/question/6073431

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