Determine the probability (as a decimal rounded to the nearest hundredth) of a normal random variable with a mean of three and a standard deviation of four realizing a value that is strictly greater than 1.9

Step-by-step explanation:

On a normal distribution table with z scores of 0 as the mean and the standard deviations going to the right and to the left, 1.9 on the normal curve where 3 is the mean falls at -.275 on the normal curve where 0 is the mean. I used the normal distribution z table for negative values to get that .39165 lies to the left of -.275, which means that 1 - .39165 of the data lies to the right.

The probability that the value is greater than 1.9 is .60835, or as a percentage, 60.835%.

you can't solve x because it is 2 linear equation and you only have one equation

the number of possible outcomes that do not show a 1 on the first spinner and show the number 4 on the second spinner is 2.

spinner 1 spinner 2

1                     1

1                     2

1                     3

1                     4

2                   1

2                   2

2                   3

2                   4   = not 1, and 4

3                   1

3                   2

3                   3

3                   4 = not 1, and 4

only 2 outcomes have passed the given condition.

step-by-step explanation: