CAN SOMEBODY HELP ME??? giving brainliest!!! AREA!
2 answers:
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Answer: Hello your question is poorly written attached below is the complete question
answer:
0.1515
Step-by-step explanation:
Number of bets = 200
mean number of losses = $0.2
Standard deviation = $2.74
<u>Determine P ( winning after 200 bets ) </u>
Average of 200 bets > 0. i.e. P( X > 0 )
considering normal distribution : μ = -0.2 , б = 2.74 , n = 200
applying Z - distribution
z = 0 - ( - 0.2 ) / ( 2.74 / √200 ) ≈ 1.03
∴ P ( z > 1.03 ) = 1 - P ( z < 1.03 )
= 1 - 0.8485 ( value gotten from z-table )
= 0.1515
Answer:
1. The chi-squared statistic = 10.36
The degrees of freedom = 17
The p-value for the test = 0.89
2. The range of the p-value from the Chi squared table = 0.75 < p-value < 0.90
Step-by-step explanation:
1. The Chi squared test is given as follows;
Therefore,
UTI No UTI % Total
Cranberry juice 8 42 84 50
Lactobacillus 19 30 61 49
Control 18 30 60 50
The chi-squared statistic is given as follows;
The chi-squared statistic = 10.36
The degrees of freedom, df = 18 - 1 = 17 since the all of the expected count have a minimum value of 18
With the aid of the calculator we find the p value as p as follows;
The p-value for the test = 0.89
2. The range of the p-value from the Chi squared table is given as follows;
0.75 < p-value < 0.90.