Answer:
2.28% probability that a person selected at random will have an IQ of 110 or higher
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:

What is the probability that a person selected at random will have an IQ of 110 or higher?
This is 1 subtracted by the pvalue of Z when X = 110. So



has a pvalue of 0.0228
2.28% probability that a person selected at random will have an IQ of 110 or higher
Answer:
Where is the rest of this Question???
Step-by-step explanation:
- Sample space = {TT, HH, TH, HT} where T is tail and H is head.
- Number of outcomes = 4
- The probability of getting two heads on tossing two coins

<u>Answer</u><u>:</u>
<u>D)</u><u> </u><u>2</u><u>5</u><u>%</u>
Hope you could understand.
If you have any query, feel free to ask.
Answer:
3x - 6
Step-by-step explanation:
Answer:
(53.3; 56.1)
Step-by-step explanation:
Given that:
Sample size, n = 41
Mean, xbar = 54.7
Standard deviation, s = 5.3
Confidence level, Zcritical at 90% = 1.645
Confidence interval :
Xbar ± Margin of error
Margin of Error = Zcritical * s/sqrt(n)
Margin of Error = 1.645 * 5.3/sqrt(41)
Margin of Error = 1.362
Lower boundary = 54.7 - 1.362 = 53.338
Upper boundary = 54.7 + 1.362 = 56.062
(53.3 ; 56.1)