Suppose a normal distribution has a mean of 98 and a standard deviation of 6. What is P(x< or = to 110)
1 answer:
Answer:
97.8%
Step-by-step explanation:
110 is 2 standard deviations above the mean (6+6 = 12)
12+98 = 110
Looking at the standard deviation curve
P(x< or = to 110) = 1 - P(x>110)
We can find the probability that x>100 by adding anything above 2 standard deviations above the curve.
P(x>110) = 2.1+.1 = 2.2%
P(x< or = to 110) = 1 - P(x>110)
= 1- 2.2%
= 1- .022
= .978
= 97.8 %
You might be interested in
Answer:
Step-by-step explanation:
Let x represent the random variable representing the scores in the exam. Given that the scores are normally distributed with a mean of 40 and a standard deviation of 5, the diagram representing the curve and the position of the mean, the mean plus or minus one standard deviation, the mean plus or minus two standard deviations, and the mean plus or minus three standard deviations is shown in the attached photo
1 standard deviation = 5
2 standard deviations = 2 × 5 = 10
3 standard deviations = 3 × 5 = 15
1 standard deviation from the mean lies between (40 - 5) and (40 + 5)
2 standard deviations from the mean lies between (40 - 10) and (40 + 10)
3 standard deviations from the mean lies between (40 - 15) and (40 + 15)
b) We would apply the probability for normal distribution which is expressed as
z = (x - µ)/σ
Where
x = sample mean
µ = population mean
σ = standard deviation
From the information given,
µ = 40
σ = 5
the probability that a randomly selected score will be greater than 50 is expressed as
P(x > 50) = 1 - P( ≤ x 50)
For x = 50,
z = (50 - 40)/5 = 2
Looking at the normal distribution table, the probability corresponding to the z score is 0.98
P(x > 50) = 1 - 0.98 = 0.02
