Assume that SAT scores are normally distributed with mean 1518 and standard deviation 325. Round your answers to 4 decimal placesa. If 100 SAT scores are randomly selected, find the probability that they have a mean less than 1500.b. If 64 SAT scores are randomly selected, find the probability that they have a mean greater than 1600c. If 25 SAT scores are randomly selected, find the probability that they have a mean between 1550 and 1575d. If 16 SAT scores are randomly selected, find the probability that they have a mean between 1440 and 1480.e. In part c and part d, why can the central limit theorem be used even though the sample size does not exceed 30?

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.

4096 bacteria

step-by-step explanation: