Answer:
12.3
Step-by-step explanation:
Step 1
We find the mean
The data list shows the scores of ten students in Mr. Smith's math class. 61, 67, 81, 83, 87, 88, 89, 90, 98, 100
Mean = Sum of terms/Number of terms
Number of terms = 10
Mean = 61 + 67 + 81 + 83 + 87 + 88 + 89 + 90 + 98 + 100/10
Mean = 844/10
Mean = 84.4
Step 2
Standard deviation
The formula for sample standard deviation =
√(x - Mean)²/n - 1
= √[(61 - 84.4)² + (67 - 84.4)² + (81 - 84.4)² + (83 - 84.4)² + (87 - 84.4)² + (88 - 84.4)² + (89 - 84.4)² + (90 - 84.4)² + (98 - 84.4)² + (100 - 84.4)²]/10 - 1
=√ 547.56 + 302.76 + 11.56 + 1.96 + 6.76 + 12.96 + 21.16 + 31.36 + 184.96 + 243.36/10 - 1
= √1364.4/9
= √151.6
= 12.31259518
Approximately to the nearest tenth = 12.3
The standard deviation = 12.3
When you make a plan on how much money you’ll waste
Answer: Git gud
Step-by-step explanation: Just do it.
The probability that you pick a striped shirt AND a blue suit is 1/4 AND a pink tie is 3/8.
According to the statement
Number of blue suit = 1
Number of black suit = 2
Number of brown suit = 1
Number of blue ties = 2
Number of red ties = 3
Number of pink ties = 3
Now we find the probability by its formula
Probability that picked suit is blue = blue suits/ total number of suits
Probability = 1/4
Now,
Probability that picked tie is pink = Pink tie / total number of ties
Probability = 3/8
So, The probability that you pick a striped shirt AND a blue suit is 1/4 AND a pink tie is 3/8.
Learn more about PROBABILITY here brainly.com/question/7965468
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Answer:
(c) The t-distribution is bell shaped
(e) The t-distribution has more values at the extremes than a standard normal distribution
(f) The t-distribution is centered at 0
Step-by-step explanation:
When the sample size is small or population variance is unknown, and we have to estimate population parameter then t- probability distribution is used.
The t-distribution has properties as follows:
- The mean of t-distribution is 0. Thus, option (f) is correct.
- t-distribution is normal. Thus, option (c) is correct.
- As compare to standard normal distribution, t-distribution has more values at the extremes. Thus, option (e) is correct.
