Answer:
a. 0.2898
b. 0.0218
c. 0.1210
d. 0.1515
e. This is because the population is normally distributed.
Step-by-step explanation:
Assume that SAT scores are normally distributed with mean 1518 and standard deviation 325. Round your answers to 4 decimal places
We are using the z score formula when random samples
This is given as:
z = (x-μ)/σ/√n
where x is the raw score
μ is the population mean
σ is the population standard deviation.
n is the random number of samples
a.If 100 SAT scores are randomly selected, find the probability that they have a mean less than 1500.
For x = 1500, n = 100
z = 1500 - 1518/325/√100
z = -18/325/10
z = -18/32.5
z = -0.55385
Probability value from Z-Table:
P(x<1500) = 0.28984
Approximately = 0.2898
b. If 64 SAT scores are randomly selected, find the probability that they have a mean greater than 1600
For x = 1600, n = 64
= z = 1600 - 1518/325/√64.
z= 1600 - 1518 /325/8
z = 2.01846
Probability value from Z-Table:
P(x<1600) = 0.97823
P(x>1600) = 1 - P(x<1600) = 0.021772
Approximately = 0.0218
c. If 25 SAT scores are randomly selected, find the probability that they have a mean between 1550 and 1575
For x = 1550, n = 25
z = 1550 - 1518/325/√25
z = 1550 - 1518/325/5
z = 1550 - 1518/65
= 0.49231
Probability value from Z-Table:
P(x = 1550) = 0.68875
For x = 1575 , n = 25
z = 1575 - 1518/325/√25
z = 1575 - 1518/325/5
z = 1575 - 1518/65
z = 0.87692
Probability value from Z-Table:
P(x=1575) = 0.80974
The probability that they have a mean between 1550 and 1575
P(x = 1575) - P(x = 1550)
= 0.80974 - 0.68875
= 0.12099
Approximately = 0.1210
d. If 16 SAT scores are randomly selected, find the probability that they have a mean between 1440 and 1480
For x = 1440, n = 16
z = 1440 - 1518/325/√16
= -0.96
Probability value from Z-Table:
P(x = 1440) = 0.16853
For x = 1480, n = 16
z = 1480 - 1518/325/√16
=-0.46769
Probability value from Z-Table:
P(x = 1480) = 0.32
The probability that they have a mean between 1440 and 1480
P(x = 1480) - P(x = 1440)
= 0.32 - 0.16853
= 0.15147
Approximately = 0.1515
e. In part c and part d, why can the central limit theorem be used even though the sample size does not exceed 30?
The central theorem can be used even though the sample size does not exceed 30 because the population is normally distributed.
. x=39.472 which is rounded to 39%
