What is the value of 3X squared +5 and X +25 when x=3
1 answer:
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Answer:
a) 0.5.
b) 0.8413
c) 0.8413
d) 0.6826
e) 0.9332
f) 1
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
(a) P(x > 6) =
This is 1 subtracted by the pvalue of Z when X = 6. So
has a pvalue of 0.5.
1 - 0.5 = 0.5.
(b) P(x < 6.2)=
This is the pvalue of Z when X = 6.2. So
has a pvalue of 0.8413
(c) P(x ≤ 6.2) =
In the normal distribution, the probability of an exact value, for example, P(X = 6.2), is always zero, which means that P(x ≤ 6.2) = P(x < 6.2) = 0.8413.
(d) P(5.8 < x < 6.2) =
This is the pvalue of Z when X = 6.2 subtracted by the pvalue of Z when X 5.8.
X = 6.2
has a pvalue of 0.8413
X = 5.8
has a pvalue of 0.1587
0.8413 - 0.1587 = 0.6826
(e) P(x > 5.7) =
This is 1 subtracted by the pvalue of Z when X = 5.7.
has a pvalue of 0.0668
1 - 0.0668 = 0.9332
(f) P(x > 5) =
This is 1 subtracted by the pvalue of Z when X = 5.
has a pvalue of 0.
1 - 0 = 1
Neptune's distance from the sun is d₁ = 4.503 x 10⁹ km.
Mercury's distance from the sun is d₂ = 5.791 x 10⁷ km.
Calculate the number of times d₁ is greater than d₂.
Answer: 7.7759 x 10⁻¹ times
Mercury's distance from the sun is d₂ = 5.791 x 10⁷ km.
Calculate the number of times d₁ is greater than d₂.
Answer: 7.7759 x 10⁻¹ times