Pls help this is so harddd for meee
1 answer:
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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 =
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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