Answer:
The probability that a household has at least one of these appliances is 0.95
Step-by-step explanation:
Percentage of households having radios P(R) = 75% = 0.75
Percentage of households having electric irons P(I) = 65% = 0.65
Percentage of households having electric toasters P(T) = 55% = 0.55
Percentage of household having iron and radio P(I∩R) = 50% = 0.5
Percentage of household having radios and toasters P(R∩T) = 40% = 0.40
Percentage of household having iron and toasters P(I∩T) = 30% = 0.30
Percentage of household having all three P(I∩R∩T) = 20% = 0.20
Probability of households having at least one of the appliance can be calculated using the rule:
P(at least one of the three) = P(R) +P(I) + P(T) - P(I∩R) - P(R∩T) - P(I∩T) + P(I∩R∩T)
P(at least one of the three)=0.75 + 0.65 + 0.55 - 0.50 - 0.40 - 0.30 + 0.20 P(at least one of the three) = 0.95
The probability that a household has at least one of these appliances is 0.95
C=7n because
28 divided 4 is 7
7 is the proportional
Answer:
CI = (98.11 , 98.49)
The value of 98.6°F suggests that this is significantly higher
Step-by-step explanation:
Data provided in the question:
sample size, n = 103
Mean temperature, μ = 98.3
°
Standard deviation, σ = 0.73
Degrees of freedom, df = n - 1 = 102
Now,
For Confidence level of 99%, and df = 102, the t-value = 2.62 [from the standard t table]
Therefore,
CI = 
Thus,
Lower limit of CI = 
or
Lower limit of CI = 
or
Lower limit of CI = 98.11
and,
Upper limit of CI = 
or
Upper limit of CI = 
or
Upper limit of CI = 98.49
Hence,
CI = (98.11 , 98.49)
The value of 98.6°F suggests that this is significantly higher and the mean temperature could very possibly be 98.6°F