Using the normal distribution, it is found that he needs a score of 605 in exam b in order to do equivalently well as he did on exam a.
The z-score of a measure X of a normally distributed variable with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
For the z-score for exam a, we have that the parameters are given as follows:
[tex]X = 22, \mu = 40, \sigma = 10[/tex]
Hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{22 - 40}{10}[/tex]
Z = -1.8.
Then, for the equivalent grade X in exam B, we have that:
[tex]Z = -1.8, \mu = 650, \sigma = 25[/tex]
Then:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-1.8 = \frac{X - 650}{25}[/tex]
X - 650 = -1.8 x 25
X = 605.
More can be learned about the normal distribution at https://brainly.com/question/24663213
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