The probability would be 0.1971.
We will calculate a z-score for each end of this interval.
z = (X-μ)/σ
For the lower limit:
z = (1100-1050)/218 = 50/218 = 0.23
For the upper limit:
z = (1225-1050)/218 = 175/218 = 0.80
Using a z-table (http://www.z-table.com) we see that the area under the curve to the left of, less than, the lower limit is 0.5910. The area under the curve to the left of, less than, the upper limit is 0.7881. To find the area between them, we subtract:
0.7881 - 0.5910 = 0.1971
Answer:
20%
Step-by-step explanation:
1/5 of 1 hour = 20 minutes
so 1/5 = 20%
25% + 20% = 45%
0.35 = 35%
45% + 35% = 80%
100% - 80% = 20%
Answer = 20%
Hope this helps!
Answer:
( √15 + 8)/7
Step-by-step explanation:
TanA = -√15
.we are to find tan(A-π/4).
In trigonometry
Tan(A-B) = TanA - TanB/1+ tanAtanB
Hence:
tan(A-π/4) = TanA - Tanπ/4/1+ tanAtanπ/4
Substitute tan A value into the formula
tan(A-π/4) = -√15-tanπ/4 / 1+(-√15)(tanπ/4
tan(A-π/4) = -√15-1/1-√15
Rationalize
-√15-1/1-√15 × 1+√15/1+√15
= -√15-√225-1-√15/(1-√225)
= -2√15-15-1/1-15
= -2√15 -16/(-14)
= -2(√15+8)/-14
= √15 + 8/7
Hence the required value is ( √15 + 8)/7
The answer to your question is 3
Using the normal distribution, it is found that 7.64% of of sample means are greater than 8.8 hours.
<h3>Normal Probability Distribution</h3>
The z-score of a measure X of a normally distributed variable with mean
and standard deviation
is given by:

- The z-score measures how many standard deviations the measure is above or below the mean.
- Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
- By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation
.
The parameters are given as follows:

The proportion of sample means greater than 8.8 hours is <u>one subtracted by the p-value of Z when X = 8.8</u>, hence:

By the Central Limit Theorem


Z = 1.43
Z = 1.43 has a p-value of 0.9236.
1 - 0.9236 = 0.0764.
7.64% of of sample means are greater than 8.8 hours.
More can be learned about the normal distribution at brainly.com/question/25800303
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